Class 10 Maths Chapter 5 AP Arithmetic Progression Solutions with MCQs

Class 10 Maths Exercise 5.1

1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.
See Solution(i) Fare for 1 km a₁ = ₹15
Fare for 2 km a₂ = ₹(15 + 8) = ₹23
Fare for 3 km a₃ = ₹(23 + 8) = ₹31
Fare for 4 km a₄ = ₹(31 + 8) = ₹39
a₂ – a₁ = 23 – 15 = 8
a₃ – a₂ = 31 – 23 = 8
a₄ – a₃ = 39 – 31 = 8
The difference between the successive terms are same. Hence, it is an A.P.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
See Solution(ii) Let, the amount of air in the cylinder a₁ = x
Amount of air left, after first removal a₂ = x – (1/4)x = (3/4)x
Amount of air left, after second removal a₃ = (3/4)x – (1/4)(3/4)x = (9/16)x
Amount of air left, after third removal a₄ = (9/16)x – (1/4)(9/16)x = (27/64)x
a₂ – a₁ = (3/4)x – x = (-1/4)x
a₃ – a₂ = (9/16)x – (3/4)x = (-3/16)x
a₄ – a₃ = (27/64)x – (9/16)x = (-9/64)x
The difference between the successive terms are not same. Hence, it is not an A.P.

(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.
See Solution(iii) Cost for digging 1 m deep a₁ = ₹150
Cost for digging 2 m deep a₂ = ₹(150 + 50) = ₹200
Cost for digging 3 m deep a₃ = ₹(200 + 50) = ₹250
Cost for digging 4 m deep a₄ = ₹(250 + 50) = ₹300
a₂ – a₁ = 200 – 150 = 50
a₃ – a₂ = 250 – 200 = 50
a₄ – a₃ = 300 – 250 = 50
The difference between the successive terms are same. Hence, it is an A.P.

(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.
See Solution(iv) Rate = 8% per annum, Principal a₁ = ₹10000
Amount after 1 year a₂ = ₹10000 (1 + 8/100)¹ = ₹10800
Amount after 2 years a₃ = ₹10800 (1 + 8/100)² = ₹12597
Amount after 3 years a₄ = ₹12597 (1 + 8/100)³ = ₹15869
a₂ – a₁ = 10800 – 10000 = 800
a₃ – a₂ = 12597 – 10800 = 1797
a₄ – a₃ = 15869 – 12597 = 3272
The difference between the successive terms are not same. Hence, it is not an A.P.

2. Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10
See Solution(i) a = 10, d = 10
First term a₁ = a = 10
Second term a₂ = a₁ + d = 10 + 10 = 20
Third term a₃ = a₂ + d = 20 + 10 = 30
Fourth term a₄ = a₃ + d = 30 + 10 = 40

(ii) a = -2, d = 0
See Solution(ii) a = -2, d = 0
First term a₁ = a = -2
Second term a₂ = a₁ + d = -2 + 0 = -2
Third term a₃ = a₂ + d = -2 + 0 = -2
Fourth term a₄ = a₃ + d = -2 + 0 = -2

(iii) a = 4, d = -3
See Solution(iii) a = 4, d = -3
First term a₁ = a = 4
Second term a₂ = a₁ + d = 4 – 3 = 1
Third term a₃ = a₂ + d = 1 – 3 = -2
Fourth term a₄ = a₃ + d = -2 – 3 = -5

(iv) a = -1, d = ½
See Solution(iv) a = -1, d = ½
First term a₁ = a = -1
Second term a₂ = a₁ + d = -1 + ½ = -½
Third term a₃ = a₂ + d = -½ + ½ = 0
Fourth term a₄ = a₃ + d = 0 + ½ = ½

(v) a = -1.25, d = -0.25
See Solution(v) a = -1.25, d = -0.25
First term a₁ = a = -1.25
Second term a₂ = a₁ + d = -1.25 – 0.25 = -1.50
Third term a₃ = a₂ + d = -1.50 – 0.25 = -1.75
Fourth term a₄ = a₃ + d = -1.75 – 0.25 = -2.00

3. For the following APs, write the first term and the common difference:
(i) 3, 1, -1, -3, …
(ii) -5, -1, 3, 7, …
(iii) 1/3, 5/3, 9/3, 13/3, …
(iv) 0.6, 1.7, 2.8, 3.9, …
See Solution(i) First term a = 3
Common difference d = a_2 – a_1 = 1 – 3 = -2
(ii) First term a = -5
Common difference d = a_2 – a_1 = -1 – (-5) = 4
(iii) First term a = 1/3
Common difference d = a_2 – a_1 = 5/3 – 1/3 = 4/3
(iv) First term a = 0.6
Common difference d = a_2 – a_1 = 1.7 – 0.6 = 1.1

4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, …
See Solution(i) 2, 4, 8, 16, …
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
a₄ – a₃ = 16 – 8 = 8
The difference between the successive terms are not same. Hence, it is not an A.P.

(ii) 2, 5/2, 3, 7/2, …
See Solution(ii) 2, 5/2, 3, 7/2, …
a₂ – a₁ = 5/2 – 2 = 5/2 – 4/2 = 1/2
a₃ – a₂ = 3 – 5/2 = 6/2 – 5/2 = 1/2
a₄ – a₃ = 7/2 – 3 = 7/2 – 6/2 = 1/2
The difference between the successive terms are same. Hence, it is an A.P.
Common difference = 1/2, next three terms of this AP is as follows:
Fifth term a₅ = a₄ + d = 7/2 + 1/2 = 8/2 = 4
Sixth term a₆ = a₅ + d = 4 + 1/2 = 8/2 + 1/2 = 9/2
Seventh term a₇ = a₆ + d = 9/2 + 1/2 = 10/2 = 5

(iii) -1.2, -3.2, -5.2, -7.2, …
See Solution(iii) -1.2, -3.2, -5.2, -7.2, …
a₂ – a₁ = -3.2 – (-1.2) = -2.0
a₃ – a₂ = -5.2 – (-3.2) = -2.0
a₄ – a₃ = -7.2 – (-5.2) = -2.0
The difference between the successive terms are same. Hence, it is an A.P.
Common difference = -2.0, next three terms of this AP is as follows:
Fifth term a₅ = a₄ + d = -7.2 – 2.0 = -9.2
Sixth term a₆ = a₅ + d = -9.2 – 2.0 = -11.2
Seventh term a₇ = a₆ + d = -11.2 – 2.0 = -13.21

(iv) -10, -6, -2, 2, …
See Solution(iv) -10, -6, -2, 2, …
a₂ – a₁ = -6 – (-10) = 4
a₃ – a₂ = -2 – (-6) = 4
a₄ – a₃ = 2 – (-2) = 4
The difference between the successive terms are same. Hence, it is an A.P.
Common difference = 4, next three terms of this AP is as follows:
Fifth term a₅ = a₄ + d = 2 + 4 = 6
Sixth term a₆ = a₅ + d = 6 + 4 = 10
Seventh term a₇ = a₆ + d = 10 + 42 = 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …
See Solution(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …
a₂ – a₁ = 3 + √2 – 3 = √2
a₃ – a₂ = 3 + 2√2 – (3 + √2) = √2
a₄ – a₃ = 3 + 3√2 – (3 + 2√2) = √2
The difference between the successive terms are same. Hence, it is an A.P.
Common difference = √2, next three terms of this AP is as follows:
Fifth term a₅ = a₄ + d = 3 + 3√2 + √2 = 3 + 4√2
Sixth term a₆ = a₅ + d = 3 + 4√2 + √2 = 3 + 5√2
Seventh term a₇ = a₆ + d = 3 + 5√2 + √2 = 3 + 6√2

(vi) 0.2, 0.22, 0.222, 0.2222, …
See Solution(vi) 0.2, 0.22, 0.222, 0.2222, …
a₂ – a₁ = 0.22 – 0.2 = 0.02
a₃ – a₂ = 0.222 – 0.22 = 0.002
a₄ – a₃ = 0.2222 – 0.222 = 0.0002
The difference between the successive terms are not same. Hence, it is not an A.P.

(vii) 0, -4, -8, -12, …
See Solution(vii) 0, -4, -8, -12, …
a₂ – a₁ = -4 – 0 = -4
a₃ – a₂ = -8 – (-4) = -4
a₄ – a₃ = -12 – (-8) = -4
The difference between the successive terms are same. Hence, it is an A.P.
Common difference = -4, next three terms of this AP is as follows:
Fifth term a₅ = a₄ + d = -12 – 4 = -16
Sixth term a₆ = a₅ + d = -16 – 4 = -20
Seventh term a₇ = a₆ + d = -20 – 4 = -24

(viii) -1/2, -1/2, -1/2, -1/2, …
See Solution a₂ – a₁ = -1/2 – (-1/2) = 0
a₃ – a₂ = -1/2 – (-1/2) = 0
a₄ – a₃ = -1/2 – (-1/2) = 0
The difference between the successive terms are same. Hence, it is an A.P.
Common difference = 0, next three terms of this AP is as follows:
Fifth term a₅ = a₄ + d = -1/2 + 0 = -1/2
Sixth term a₆ = a₅ + d = -1/2 + 0 = -1/2
Seventh term a₇ = a₆ + d = -1/2 + 0 = -1/2

(ix) 1, 3, 9, 27, …
See Solution(ix) 1, 3, 9, 27, …
a₂ – a₁ = 3 – 1 = 2
a₃ – a₂ = 9 – 3 = 6
a₄ – a₃ = 27 – 9 = 18
The difference between the successive terms are not same. Hence, it is not an A.P.

(x) a, 2a, 3a, 4a, …
See Solution(x) a, 2a, 3a, 4a, …
a₂ – a₁ = 2a – a = a
a₃ – a₂ = 3a – 2a = a
a₄ – a₃ = 4a – 3a = a
The difference between the successive terms are same. Hence, it is an A.P.
Common difference = a, next three terms of this AP is as follows:
Fifth term a₅ = a₄ + d = 4a + a = 5a
Sixth term a₆ = a₅ + d = 5a + a = 6a
Seventh term a₇ = a₆ + d = 6a + a = 7a

(xi) a, a², a³, a⁴, …
See Solution(xi) a, a², a³, a⁴, …
a₂ – a₁ = a² – a
a₃ – a₂ = a³ – a²
a₄ – a₃ = a⁴ – a³
The difference between the successive terms are not same. Hence, it is not an A.P.

(xii) √2, √8, √18, √32, …
See Solution(xii) √2, √8, √18, √32, …
a₂ – a₁ = √8 – √2 = 2√2 – √2 = √2
a₃ – a₂ = √18 – √8 = 3√2 – 2√2 = √2
a₄ – a₃ = √32 – √18 = 4√2 – 3√2 = √2
The difference between the successive terms are same. Hence, it is an A.P.
Common difference = √2, next three terms of this AP is as follows:
Fifth term a₅ = a₄ + d = √32 + √2 = 4√2 + √2 = 5√2 = √50
Sixth term a₆ = a₅ + d = 5√2 + √2 = 6√2 = √72
Seventh term a₇ = a₆ + d = 6√2 + √2 = 7√2 = √98

(xiii) √3, √6, √9, √12, …
See Solution(xiii) √3, √6, √9, √12, …
a₂ – a₁ = √6 – √3
a₃ – a₂ = √9 – √6 = 3 – √6
a₄ – a₃ = √12 – √9 = 2√3 – 3
The difference between the successive terms are not same. Hence, it is not an A.P.

(xiv) 1², 3², 5², 7², …
See Solution(xiv) 1², 3², 5², 7², …
a₂ – a₁ = 3² – 1² = 9 – 1 = 8
a₃ – a₂ = 5² – 3² = 25 – 9 = 16
a₄ – a₃ = 7² – 5² = 49 – 25 = 24
The difference between the successive terms are not same. Hence, it is not an A.P.

(xv) 1², 5², 7², 73, …
See Solution(xv) 1², 5², 7², 73, …
a₂ – a₁ = 5² – 1² = 25 – 1 = 24
a₃ – a₂ = 7² – 5² = 49 – 25 = 24
a₄ – a₃ = 73 – 7² = 73 – 49 = 24
The difference between the successive terms are same. Hence, it is an A.P.
Common difference = 24, next three terms of this AP is as follows:
Fifth term a₅ = a₄ + d = 73 + 24 = 97
Sixth term a₆ = a₅ + d = 97 + 24 = 121
Seventh term a₇ = a₆ + d = 121 + 24 = 145

Class 10 Maths Exercise 5.2

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and aₙ the nth term of the AP:
See Solution(i) Here, a = 7, d = 3 and n = 8, so, putting the values in aₙ = a + (n – 1)d, we get:
aₙ = 7 + (8 – 1)(3) ⇒ aₙ = 7 + 21 = 28
(ii) Here, a = -18, n = 10 and aₙ = 0, so, putting the values in aₙ = a + (n – 1)d, we get:
0 = -18 + (10 – 1)d ⇒ 18 = 9d ⇒ d = 2
(iii) Here, d = -3, n = 18 and aₙ = -5, so, putting the values in aₙ = a + (n – 1)d, we get:
-5 = a + (18 – 1)(-3) ⇒ -5 = a – 51 ⇒ a = 46
(iv) Here, a = -18.9, d = 2.5 and aₙ = 3.6, putting the values in aₙ = a + (n – 1)d, we get:
3.6 = -18.9 + (n – 1)(2.5) ⇒ 3.6 = -18.9 + 2.5n – 2.5 ⇒ 2.5n = 3.6 + 21.4 = 25.0 ⇒ n = 10
(v) Here, a = 3.5, d = 0 and n = 105, putting the values in aₙ = a + (n – 1)d, we get:
aₙ = 3.5 + (105 – 1)(0) ⇒ aₙ = 3.5 + 0 = 3.5

2. Choose the correct choice in the following and justify:
(i) 30ᵗʰ term of the A.P.: 10, 7, 4, …, is
[A] 97 [B] 77 [C] −77 [D] −87
(ii) 11ᵗʰ term of the A.P.: −3, −½, 2, …, is
[A] 28 [B] 22 [C] −38 [D] −48½
See Solution(i) Here, a = 10, d = 7 − 10 = −3 and n = 30.
Therefore, putting the values in aₙ = a + (n − 1)d, we get
a₃₀ = 10 + (30 − 1)(−3)
⇒ a₃₀ = 10 − 87 = −77
Hence, the option (C) is correct.
(ii) Here, a = −3, d = −½ − (−3) = ⁵⁄₂ and n = 11.
Therefore, putting the values in aₙ = a + (n − 1)d, we get
a₁₁ = −3 + (11 − 1)(⁵⁄₂)
⇒ a₁₁ = −3 + 25 = 22
Hence, the option (B) is correct.

3. In the following APs, find the missing terms in the boxes:
(i) 2, __, 26
See Solution(i) Here, a = 2 and a₃ = 26. To find: a₂
Given that: a₃ = a + (3 – 1)×d = 26
=> 2 + 2×d = 26
=> 2*d = 24
=> d = 12
Therefore, a₂ = a + (2 – 1)×d = 2 + 12 = 14
The missing term is 14.

(ii) __, 13, __, 3
See Solution(ii) Here, a₂ = 13 and a₄ = 3. To find: a₁ and a₃
Given that: a₂ = a + (2 – 1) × d = 13
=> a + d = 13 … (1)
=> a = 13 – d
and a₄ = 3
=> a + (4 – 1)×d = 3
=> a + 3×d = 3
Putting the value of a from equation (1), we get
(13 – d) + 3×d = 3
=> 13 + 2×d = 3
=> 2×d = 3 – 13
=> 2×d = -10
=> d = -5
Putting the value of d in equation (1), we get
a = 13 – (-5) = 13 + 5 = 18
Therefore, a₁ = a + (1 – 1)×d = 18 + 0×(-5) = 18
and a₃ = a + (3 – 1) × d = 18 + 2 × (-5) = 18 – 10 = 8
The missing terms are 18 and 8.

(iii) 5, __, __, 9 1/2
See SolutionHere, a = 5 and a_4 = 9 1/2. To find: a₂ and a₃
Given that: a_4 = a + (4-1)d = 9 1/2
=> 5 + 3d = 19/2
=> 3d = 19/2 – 5 = 9/2
=> d = 3/2
Therefore, a_2 = a + d = 5 + 3/2 = 6 1/2
and a_3 = a + 2d = 5 + 2(3/2) = 8

(iv) -4, __, __, __, __, 6
See Solution(iv) Here, a = -4 and a₆ = 6. To find: a₂, a₃, a₄ and a₅
Given that: a₆ = a + (6 – 1)d = 6 ⇒ -4 + 5d = 6
⇒ 5d = 10 ⇒ d = 2
Therefore, a₂ = a + d = -4 + 2 = -2
a₃ = a + 2d = -4 + 2(2) = 0
a₄ = a + 3d = -4 + 3(2) = 2
a₅ = a + 4d = -4 + 4(2) = 4

(v) __, 38, __, __, __, -22
See Solution(v) Here, a₂ = 38 and a₆ = -22.
To find: a₁, a₃, a₄ and a₅
Given that: a₂ = a + (2 – 1)d = 38
⇒ a + d = 38
⇒ a = 38 – d … (1)
and a₆ = -22 ⇒ a + 5d = -22
Putting the value of a from equation (1), we get
38 – d + 5d = -22 ⇒ 38 + 4d = -22 ⇒ 4d = -60 ⇒ d = -15
Putting the value of d in equation (1), we get
a = 38 – (-15) = 53
Therefore, a₁ = a = 53
a₃ = a + 2d = 53 + 2(-15) = 53 – 30 = 23
a₄ = a + 3d = 53 + 3(-15) = 53 – 45 = 8
a₅ = a + 4d = 53 + 4(-15) = 53 – 60 = -7

4. Which term of the A.P.: 3, 8, 13, 18, … is 78?
See SolutionHere, a = 3 and d = 8 – 3 = 5.
Let, nth term of the A.P. is 78.
Therefore, a_n = 78
=> a + (n – 1)d = 78
=> 3 + (n – 1)(5) = 78
=> (n – 1)(5) = 75
=> n – 1 = 15
=> n = 16
Hence, 16th term of the A.P.: 3, 8, 13, 18, … is 78.

5. Find the number of terms in each of the following APs:
(i) 7, 13, 19, …, 205
See Solution(i) Here, a = 7 and d = 13 – 7 = 6.
Let the total number of terms in the AP be n.
Therefore, a_n = 205
=> a + (n – 1)d = 205
=> 7 + (n – 1)(6) = 205
=> (n – 1)(6) = 198
=> n – 1 = 33
=> n = 34
Hence, there are 34 terms in the given AP.

(ii) 18, 15 1/2, 13, …, -47
See Solution(ii) Here, a = 18 and d = 15 1/2 – 18 = 31/2 – 36/2 = -5/2.
Let the total number of terms in the AP be n.
Therefore, a_n = -47
=> a + (n – 1)d = -47
=> 18 + (n – 1)(-5/2) = -47
=> (n – 1)(-5/2) = -47 – 18
=> (n – 1)(-5/2) = -65
=> n – 1 = -65 × (-2/5)
=> n – 1 = (65 × 2) / 5
=> n – 1 = 13 × 2
=> n – 1 = 26
=> n = 27
Hence, there are 27 terms in the given AP.

6. Check whether -150 is a term of the AP: 11, 8, 5, 2…
See SolutionHere, a = 11 and d = 8 – 11 = -3.
Let, the nth term of the A.P. is -150.
Therefore, a_n = -150
=> a + (n – 1)d = -150
=> 11 + (n – 1)(-3) = -150
=> 11 – 3n + 3 = -150
=> 14 – 3n = -150
=> -3n = -150 – 14
=> -3n = -164
=> n = 164 / 3
=> n = 54 2/3
Here, n is not a natural number, therefore, -150 is not the term of A.P. 11, 8, 5, 2, …

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
See SolutionHere, a₁₁ = 38 and a₁₆ = 73. To find: a₃₁
Given that: a₁₁ = a + (11 – 1)d = 38
⇒ a + 10d = 38
⇒ a = 38 – 10d … (1)
and a₁₆ = 73 ⇒ a + 15d = 73
Putting the value of a from equation (1), we get
38 – 10d + 15d = 73
⇒ 38 + 5d = 73
⇒ 5d = 35 ⇒ d = 7
Putting the value of d in equation (1), we get
a = 38 – 10(7) = -32
Therefore, a₃₁ = a + 30d = -32 + 30(7) = 178
Hence, the 31st term is 178.

8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
See SolutionHere, a₃ = 12 and a₅₀ = 106. To find: a₂₉
Given that: a₃ = a + (3 – 1)d = 12
⇒ a + 2d = 12
⇒ a = 12 – 2d … (1)
and a₅₀ = 106
⇒ a + 49d = 106
Putting the value of a from equation (1), we get
12 – 2d + 49d = 106
⇒ 12 + 47d = 106
⇒ 47d = 94 ⇒ d = 2
Putting the value of d in equation (1), we get
a = 12 – 2(2) = 8
Therefore, a₂₉ = a + 28d = 8 + 28(2) = 64
Hence, the 29th term of the AP is 64.

9. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
See SolutionHere, a₃ = 4 and a₉ = -8. To find: n, where aₙ = 0.
Given that: a₃ = a + (3 – 1)d = 4
⇒ a + 2d = 4
⇒ a = 4 – 2d … (1)
and a₉ = -8
⇒ a + 8d = -8
Putting the value of a from equation (1), we get
4 – 2d + 8d = -8
⇒ 4 + 6d = -8
⇒ 6d = -12 ⇒ d = -2
Putting the value of d in equation (1), we get
a = 4 – 2(-2) = 8
Putting the values in aₙ = 0, we get: aₙ = a + (n – 1)d = 0
⇒ 8 + (n – 1)(-2) = 0
⇒ 8 – 2(n – 1) = 0
⇒ 8 – 2n + 2 = 0
⇒ 10 – 2n = 0
⇒ -2n = -10
⇒ n = 5
Hence, the 5th term of this AP is zero.

10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
See SolutionLet the first term = a and common difference = d
According to question, a₁₇ = a₁₀ + 7
⇒ a + 16d = a + 9d + 7
⇒ 7d = 7 ⇒ d = 1
Hence, the common difference is 1.

11. Which term of the AP: 3, 15, 27, 39 . . . will be 132 more than its 54th term?
See SolutionFirst term = 3 and common difference = 15 – 3 = 12
Let the nth term of AP: 3, 15, 27, 39 . . . will be 132 more than its 54th term.
Therefore, aₙ = a₅₄ + 132
⇒ a + (n – 1)d = a + 53d + 132
⇒ (n – 1)(12) = 53 × 12 + 132
⇒ (n – 1)(12) = 768
⇒ n – 1 = 768/12 = 64
⇒ n = 65
Hence, 65ᵗʰ term of the AP: 3, 15, 27, 39 . . . will be 132 more than its 54th term.

12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
See SolutionLet the first term of the first AP = A and the common difference = d
Let the first term of the second AP = a and the common difference = d
Difference between their 100ᵗʰ term = A₁₀₀ – a₁₀₀ = 100
⇒ (A + 99d) – (a + 99d) = 100
⇒ A – a = 100
Difference between their 1000ᵗʰ term = A₁₀₀₀ – a₁₀₀₀
= (A + 999d) – (a + 999d)
= A – a = 100 [∵ A – a = 100]
Hence, the difference between their 1000th terms is 100.

13. How many three-digit numbers are divisible by 7?
See SolutionThree digit numbers divisible by 7: 105, 112, 119, …, 994
Let the total number of these numbers be n.
Here, a = 105 and d = 112 – 105 = 7.
To find: n, where aₙ = 994.
Given that: aₙ = a + (n – 1)d = 994
⇒ 105 + (n – 1)(7) = 994
⇒ 7(n – 1) = 889 ⇒ n – 1 = 889/7 = 127
⇒ n = 128
Hence, there are 128 three digits numbers which are divisible by 7.

14. How many multiples of 4 lie between 10 and 250?
See SolutionMultiples of 4 lie between 10 and 250: 12, 16, 20, …, 248
Let the total number of multiples of 4 lie between 10 and 250 be n.
Here, a = 12 and d = 16 – 12 = 4.
To find: n, where aₙ = 248.
Given that: aₙ = a + (n – 1)d = 248
⇒ 12 + (n – 1)(4) = 248
⇒ 4(n – 1) = 236
⇒ n – 1 = 236/4 = 59
⇒ n = 60
Hence, the total number of multiples of 4 lie between 10 and 250 is 60.

15. For what value of n, are the nth terms of two APs: 63, 65, 67 . . . and 3, 10, 17 . . . equal?
See SolutionFirst term of first AP = A = 63 and common difference = D = 65 – 63 = 2
Therefore, Aₙ = A + (n – 1)D ⇒ Aₙ = 63 + (n – 1)2
First term of second AP = a = 3 and common difference = d = 10 – 3 = 7
Therefore, aₙ = a + (n – 1)d
⇒ aₙ = 3 + (n – 1)7
According to question, Aₙ = aₙ
⇒ 63 + (n – 1)2 = 3 + (n – 1)7
⇒ 63 + 2n – 2 = 3 + 7n – 7
⇒ 63 + 2n – 2 = 3 + 7n – 7
⇒ 65 = 5n
⇒ n = 13
Hence, the 13ᵗʰ term of both the APs are equal.

16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
See SolutionLet the first term of the AP = a and common difference = d
Third term = 16
⇒ a₃ = 16 ⇒ a + 2d = 16 … (1)
7th term exceeds the 5th term by 12, therefore a₇ = a₅ + 12
⇒ a + 6d = a + 4d + 12
⇒ 2d = 12
⇒ d = 6
Putting the value of d in equation (1), we get a + 2(6) = 16
⇒ a = 4
Hence, the A.P. = a, a + d, a + 2d, … = 4, 10, 16, …

17. Find the 20th term from the last term of the AP: 3, 8, 13… 253.
See SolutionThe 20th term from the last term of the AP: 3, 8, 13… 253 = the 20th term from the beginning of the AP: 253, …, 13, 8, 3.
In the A.P.: 253, …, 13, 8, 3, first term = 253 and common difference = 3 – 8 = -5
Therefore, a₂₀ = a + 19d
⇒ a₂₀ = 253 + 19(-5) = 253 – 95 = 158

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
See SolutionLet the first term of the AP = a and common difference = d
According to first condition, a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12 …(1)
According to second condition, a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22
Putting the value of a from equation (1), we get
(12 – 5d) + 7d = 22
2d = 10
d = 5
Putting the value of d in equation (1), we get a = 12 – 5(5) = -13
The first three terms of this AP: a, a + d, a + 2d = -13, -8, -3.

19. Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
See SolutionStarting salary = a = ₹5000 annual increment (common difference) = d = ₹200
Let, after n years, his salary become ₹7000.
Therefore, an = 7000
a + (n-1)d = 7000
5000 + (n-1)(200) = 7000
(n-1)(200) = 2000
n-1 = 10
n = 11
Hence, in 11th year his salary become ₹7000.

20. Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.
See SolutionSavings for the first week = a = ₹5 increment in savings = d = ₹1.75
Let, her saving become ₹20.75 after n weeks.
Therefore, an = 20.75
a + (n-1)d = 20.75
5 + (n-1)(1.75) = 20.75
(n-1)(1.75) = 15.75
n-1 = 15.75 / 1.75
n-1 = 9
n = 10
Hence, her saving become ₹20.75, after 10 weeks.

Class 10 Maths Exercise 5.3

1. Find the sum of the following APs:
(i) 2, 7, 12, …, to 10 terms
See Solution(i) A.P.: 2, 7, 12,…
Here, a = 2 and d = 7 – 2 = 5.
The sum of n terms of an AP is given by
Sₙ = n⁄2[2a + (n-1)d]
=> S₁₀ = 10⁄2[2(2) + (10 – 1)(5)]
=> S₁₀ = 5[4 + 45] = 245

(ii) -37, -33, -29, …, to 12 terms
See Solution(ii) A.P.: -37, -33, -29,…
Here, a = -37 and d = -33 – (-37) = 4.
The sum of n terms of an AP is given by
Sₙ = n⁄2[2a + (n-1)d]
=> S₁₂ = 12⁄2[2(-37) + (12 – 1)(4)]
=> S₁₂ = 6[-74 + 44] = -180

(iii) 0.6, 1.7, 2.8, …, to 100 terms
See Solution(iii) A.P.: 0.6, 1.7, 2.8,…
Here, a = 0.6 and d = 1.7 – 0.6 = 1.1.
The sum of n terms of an AP is given by
Sₙ = n⁄2[2a + (n-1)d]
=> S₁₀₀ = 100⁄₂[2(0.6) + (100 – 1)(1.1)]
=> S₁₀₀ = 50[1.2 + 99 × 1.1] = 50[110.1] = 5505

(iv) 1/15, 1/12, 1/10, …, to 11 terms
See Solution(iv) A.P.: 1/15, 1/12, 1/10,…
Here, a = 1/15 and d = 1/12 – 1/15 = 1/60.
The sum of n terms of an AP is given by
Sₙ = n⁄₂[2a + (n-1)d]
=> S₁₁ = 11⁄2[2(1/15) + (11 – 1)(1/60)]
=> S₁₁ = 11⁄2[2/15 + 10/60]
=> S₁₁ = 11⁄2[2/15 + 1/6]
=> S₁₁ = 11⁄2[9/30] = 33/20

2. Find the sums given below:
(i) 7 + 10 1/2 + 14 + … + 84
See Solution(i) Here, a = 7 and d = 10 1/2 – 7 = 21/2 – 7 = 7/2.
Let the nth term of the A.P. is 84, therefore, aₙ = 84
=> a + (n – 1)d = 84 => 7 + (n – 1)(7/2) = 84 => (n – 1)(7/2) = 77
=> n – 1 = 22 => n = 23
The sum of n terms of an AP is given by
Sₙ = n⁄2[a + l] => S₂₃ = 23⁄₂[7 + 84]
=> S₂₃ = 23⁄2[91] = 2093/2 = 1046 1/2

(ii) 34 + 32 + 30 + … + 10
See Solution(ii) Here, a = 34 and d = 32 – 34 = -2.
Let, the nth term of the A.P. is 10.
Therefore, aₙ = 10
=> a + (n – 1)d = 10
=> 34 + (n – 1)(-2) = 10
=> (n – 1)(-2) = -24
=> n – 1 = 12
=> n = 13
The sum of n terms of an AP is given by
Sₙ = n⁄2[a + l]
=> S₁₃ = 13⁄2[34 + 10]
=> S₁₃ = 13⁄2[44]
= 13 × 22 = 286

(iii) -5 + (-8) + (-11) + … + (-230)
See Solution(iii) Here, a = -5 and d = -8 – (-5) = -3.
Let, the nth term of the A.P. is -230.
Therefore, aₙ = -230
=> a + (n – 1)d = -230
=> -5 + (n – 1)(-3) = -230
=> (n – 1)(-3) = -225
=> n – 1 = 75 => n = 76
The sum of n terms of an AP is given by
Sₙ = n⁄₂[a + l]
=> S₇₆ = 76⁄2[-5 – 230]
=> S₇₆ = 76⁄2[-235] = 38 × (-235) = -8930

3. In an AP:
(i) given a = 5, d = 3, aₙ = 50, find n and Sₙ.
See Solution(i) Here, a = 5, d = 3 and aₙ = 50.
The nth term of the A.P. is 50.
Therefore, aₙ = 50
⇒ a + (n – 1)d = 50
⇒ 5 + (n – 1)(3) = 50
⇒ (n – 1)(3) = 45
⇒ n – 1 = 15
⇒ n = 16
The sum of n terms of an AP is given by Sₙ = (n/2)[a + l]
⇒ S₁₆ = (16/2)[5 + 50]
⇒ S₁₆ = 8[55] = 440

(ii) given a = 7, a₁₃ = 35, find d and S₁₃.
See Solution(ii) Here, a = 7 and a₁₃ = 35.
The 13th term of the A.P. is 35.
Therefore, a₁₃ = 35
⇒ a + (13 − 1)d = 35
⇒ 7 + 12d = 35
⇒ 12d = 28
⇒ d = 2812=73
The sum of n terms of an AP is given by Sₙ = n/2[a+l]
⇒ S₁₃ = 132[7 + 35]
⇒ S₁₃ = 132[42] = 273

(iii) given a₁₂ = 37, d = 3, find a and S₁₂.
See SolutionHere, d = 3 and a₁₂ = 37.
The 12th term of the A.P. is 37.
Therefore, a₁₂ = 37
⇒ a + (12−1)d = 37
⇒ a + 11(3) = 37
⇒ a = 37 − 33 = 4
The sum of n terms of an AP is given by Sₙ = (n/2)[a + l]
⇒ S₁₂ = (12/2)[4 + 37]
⇒ S₁₂ = 6[41] = 246

(iv) given a₃ = 15, S₁₀ = 125, find d and a₁₀.
See SolutionHere, a₃ = 15 and S₁₀ = 125.
a₃ = 15 ⇒ a + (3−1)d = 15
⇒ a + 2d = 15
⇒ a = 15 − 2d … (1)
The sum of n terms of an AP is given by Sₙ = (n/2)[2a + (n−1)d]
⇒ S₁₀ = (10/2)[2a + (10−1)d]
⇒ 125 = 5[2a + 9d]
⇒ 2a + 9d = 25
Putting the value of a from equation (1), we get, 2(15 − 2d) + 9d = 25
⇒ 30 − 4d + 9d = 25
⇒ 5d = −5 ⇒ d = −1
Putting the value of d in equation (1), we get, ⇒ a = 15 − 2(−1) = 17
aₙ = a + (n−1)d ⇒ a₁₀ = 17 + (10−1)(−1) = 17 − 9 = 8
⇒ a₁₀ = 8

(v) given d = 5, S₉ = 75, find a and a₉.
See SolutionHere, d = 5 and S₉ = 75.
The sum of n terms of an AP is given by Sₙ = (n/2)[2a + (n−1)d]
⇒ S₉ = (9/2)[2a + (9−1)(5)] ⇒ 75 = (9/2)[2a + 40]
⇒ 75 = 9a + 180 ⇒ a = −105/9 = −35/3
aₙ = a + (n−1)d
⇒ a₉ = −35/3 + (9−1)(5) = −35/3 + 40 = 85/3
⇒ a₉ = 85/3

(vi) given a = 2, d = 8, Sn = 90, find n and an.
See SolutionHere, a = 2, d = 8 and Sₙ = 90.
The sum of n terms of an AP is given by
Sₙ = (n/2)[2a + (n−1)d]
⇒ 90 = (n/2)[2(2) + (n−1)(8)]
⇒ 90 = (n/2)[4 + 8n − 8] ⇒ 90 = (n/2)[8n − 4]
⇒ 90 = 4n² − 2n
⇒ 2n² − n − 45 = 0
⇒ 2n² − 10n + 9n − 45 = 0
⇒ 2n(n − 5) + 9(n − 5) = 0
⇒ (2n + 9)(n − 5) = 0
⇒ n − 5 = 0 [∵ 2n + 9 ≠ 0 as n ≠ −9/2]
⇒ n = 5
aₙ = a + (n−1)d
⇒ a₅ = 2 + (5−1)(8) = 2 + 32 = 34
⇒ a₅ = 34

(vii) given a = 8, an = 62, Sn = 210, find n and d.
See SolutionHere, a = 8, aₙ = 62 and Sₙ = 210.
The sum of n terms of an AP is given by Sₙ = (n/2)[a + aₙ]
⇒ 210 = (n/2)[8 + 62]
⇒ 210 = 35n
⇒ n = 210/35 = 6
aₙ = a + (n−1)d
⇒ 62 = 8 + (6−1)d
⇒ 54 = 5d ⇒ d = 54/5

(viii) given an = 4, d = 2, Sn = –14, find n and a.
See SolutionHere, aₙ = 4, d = 2 and Sₙ = −14.
aₙ = a + (n−1)d
⇒ 4 = a + (n−1)(2)
⇒ 4 = a + 2n − 2
⇒ a = 6 − 2n … (1)
The sum of n terms of an AP is given by
Sₙ = (n/2)[2a + (n−1)d]
⇒ −14 = (n/2)[2a + (n−1)(2)]
⇒ −14 = n[a + n − 1]
Putting the value of a from equation (1), we get, −14 = n[6 − 2n + n − 1]
⇒ −14 = n[5 − n]
⇒ −14 = 5n − n²
⇒ n² − 5n − 14 = 0
⇒ n² − 7n + 2n − 14 = 0
⇒ n(n − 7) + 2(n − 7) = 0
⇒ (n − 7)(n + 2) = 0
⇒ n − 7 = 0 [∵ n + 2 ≠ 0 as n ≠ −2]
⇒ n = 7
Putting the value of d in equation (1), we get
⇒ a = 6 − 2(7) = −8

(ix) given a = 3, n = 8, S = 192, find d.
See SolutionHere, a = 3, n = 8 and S = 192.
The sum of n terms of an AP is given by
Sₙ = (n/2)[a + aₙ] ⇒ 192 = (8/2)[3 + aₙ]
⇒ 192 = 4[3 + aₙ]
⇒ 3 + aₙ = 192/4 = 48
⇒ aₙ = 45 aₙ = a + (n−1)d
⇒ 45 = 3 + (8−1)d
⇒ 42 = 7d
⇒ d = 6

(x) given l = 28, S = 144, and there are total 9 terms. Find a.
See SolutionHere, l = 28, S = 144 and n = 9.
The sum of n terms of an AP is given by
Sₙ = (n/2)[a + l]
⇒ 144 = (9/2)[a + 28]
⇒ 144 × 2/9 = a + 28
⇒ 32 = a + 28 ⇒ a = 4

4. How many terms of the AP: 9, 17, 25 . . . must be taken to give a sum of 636?
See SolutionHere, a = 9, d = 17 − 9 = 8 and Sₙ = 636.
The sum of n terms of an AP is given by
Sₙ = (n/2)[2a + (n−1)d]
⇒ 636 = (n/2)[2(9) + (n−1)(8)] ⇒ 636 = n[9 + 4n − 4]
⇒ 4n² + 5n − 636 = 0 ⇒ 4n² + 53n − 48n − 636 = 0
⇒ n(4n + 53) − 12(4n + 53) = 0 ⇒ (n − 12)(4n + 53) = 0
⇒ n − 12 = 0 [∵ 4n + 53 ≠ 0 as n ≠ −53/4]
⇒ n = 12
Hence, 12 terms of the AP: 9, 17, 25 . . . must be taken to get the sum 636.

5. The first term of an AP is 5, the last term s 45 and the sum is 400. Find the number of terms and the common difference.
See SolutionHere, a = 5, aₙ = 45 and Sₙ = 400.
The sum of n terms of an AP is given by
Sₙ = (n/2)[a + aₙ]
⇒ 400 = (n/2)[5 + 45] ⇒ 400 = 25n ⇒ n = 400/25 = 16
aₙ = a + (n−1)d ⇒ 45 = 5 + (16−1)d ⇒ 40 = 15d
⇒ d = 40/15 = 8/3
Hence, the number of terms are 16 and the common difference is 8/3.

6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
See SolutionHere, a = 17, aₙ = 350 and d = 9.
aₙ = a + (n−1)d
⇒ 350 = 17 + (n−1)9
⇒ 350 = 8 + 9n
⇒ n = 342/9 = 38
The sum of n terms of an AP is given by
Sₙ = (n/2)[a + aₙ]
⇒ S₃₈ = (38/2)[17 + 350]
⇒ S₃₈ = 19 × 367
= 6973
Hence, there are 38 terms and their sum is 6973.

7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
See SolutionHere, d = 7, aₙ = 149 and n = 22.
aₙ = a + (n−1)d
⇒ 149 = a + (22−1)(7)
⇒ 149 = a + 147 ⇒ a = 2
The sum of n terms of an AP is given by
Sₙ = (n/2)[a + aₙ]
⇒ S₂₂ = (22/2)[2 + 149]
⇒ S₂₂ = 11 × 151
= 1661
Hence, the sum of first 22 terms of this AP is 1661.

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
See SolutionHere, a₂ = 14, a₃ = 18 and n = 51.
aₙ = a + (n−1)d
⇒ a₂ = a + (2−1)d ⇒ 14 = a + d
⇒ a = 14 − d … (1)
and a₃ = a + (3−1)d ⇒ 18 = a + 2d
Putting the value of a from equation (1), we get 18 = 14 − d + 2d ⇒ d = 4
Putting the value of d in equation (1), we get a = 14 − 4 = 10
The sum of n terms of an AP is given by
Sₙ = (n/2)[2a + (n−1)d]
⇒ S₅₁ = (51/2)[2(10) + (51−1)(4)]
⇒ S₅₁ = (51/2)[220] = 5610

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
See SolutionHere, S₇ = 49 and S₁₇ = 289.
The sum of n terms of an AP is given by
Sₙ = (n/2)[2a + (n−1)d]
⇒ S₇ = (7/2)[2a + (7−1)d]
⇒ 49 = (7/2)[2a + 6d]
⇒ 49 = 7(a + 3d)
⇒ 7 = a + 3d
⇒ a = 7 − 3d … (1)
and S₁₇ = (17/2)[2a + (17−1)d]
⇒ 289 = (17/2)[2a + 16d]
⇒ 289 = 17(a + 8d)
⇒ 17 = a + 8d
Putting the value of a from equation (1), we get
⇒ 17 = 7 − 3d + 8d
⇒ 5d = 10
⇒ d = 2
Putting the value of d in equation (1), we get
⇒ a = 7 − 3 × 2 = 1
The sum of n terms of an AP is given by
Sₙ = (n/2)[2a + (n−1)d]
= (n/2)[2(1) + (n−1)(2)]
= (n/2)[2 + 2n − 2] = n²

10. Show that a₁, a₂, … , aₙ, … form an AP where aₙ is defined as below:
Also find the sum of the first 15 terms in each case.
(i) aₙ = 3 + 4n
See Solutionaₙ = 3 + 4n
Putting n = 1, we get
a₁ = 3 + 4(1) = 7
Putting n = 2, we get
a₂ = 3 + 4(2) = 11
Similarly, a₃ = 3 + 4(3) = 15
a₄ = 3 + 4(4) = 19
Difference between the successive terms:
a₂ − a₁ = 11 − 7 = 4
a₃ − a₂ = 15 − 11 = 4
a₄ − a₃ = 19 − 15 = 4
The difference between successive terms are same, hence a₁, a₂, … , aₙ, … is an A.P.
The sum of n terms of an AP is given by
Sₙ = (n/2)[2a + (n−1)d]
⇒ S₁₅ = (15/2)[2(7) + (15−1)(4)]
⇒ S₁₅ = (15/2)[70] = 525

(ii) aₙ = 9 − 5n
See Solutionaₙ = 9 − 5n
Putting n = 1, we get, a₁ = 9 − 5(1) = 4
Putting n = 2, we get
a₂ = 9 − 5(2) = −1
Similarly, a₃ = 9 − 5(3) = −6
a₄ = 9 − 5(4) = −11
Difference between the successive terms:
a₂ − a₁ = −1 − 4 = −5
a₃ − a₂ = −6 − (−1) = −5
a₄ − a₃ = −11 − (−6) = −5
The difference between successive terms are same, hence a₁, a₂, … , aₙ, … is an A.P.
The sum of n terms of an AP is given by
Sₙ = (n/2)[2a + (n−1)d]
⇒ S₁₅ = (15/2)[2(4) + (15−1)(−5)] ⇒ S₁₅ = (15/2)[−62] = −465

11. If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
See SolutionThe sum of n terms of an AP is given by
Sₙ = 4n − n²
Putting n = 1, we get, first term = a₁ = S₁ = 4(1) − (1)² = 3
Putting n = 2, we get
Sum of two terms = a₁ + a₂ = S₂ = 4(2) − (2)² = 4 ⇒ a₁ + a₂ = 4
⇒ 3 + a₂ = 4 [∵ the first term a₁ = 3]
⇒ a₂ = 1
Hence, the second term is 1.
Common difference d = a₂ − a₁ = 1 − 3 = −2
Therefore, the tenth term = a₁₀ = a + 9d = 3 + 9(−2) = −16
Similarly, the nth term = aₙ = a + (n−1)d = 3 + (n−1)(−2) = 5 − 2n

12. Find the sum of the first 40 positive integers divisible by 6.
See SolutionThe first 40 positive integers divisible by 6 are 6, 12, 18, … , 240.
Here, a = 6, d = 12 − 6 = 6 and n = 40.
The sum of n terms of an AP is given by
Sₙ = (n/2)[2a + (n−1)d]
⇒ S₄₀ = (40/2)[2(6) + (40−1)(6)] = 20[12 + 234] = 20(246) = 4920
Hence, the sum of the first 40 positive integers divisible by 6 is 4920.

13. Find the sum of the first 15 multiples of 8.
See SolutionThe first 15 multiples of 8 are 8, 16, 24, … ,120.
Here, a = 8, d = 16 − 8 = 8 and n = 15.
The sum of n terms of an AP is given by
Sₙ = (n/2)[2a + (n−1)d]
⇒ S₁₅ = (15/2)[2(8) + (15−1)(8)]
= (15/2)[16 + 112]
= (15/2)(128)
= 960
Hence, the sum of the first 15 multiples of 8 is 960.

14. Find the sum of the odd numbers between 0 and 50.
See SolutionThe odd numbers between 0 and 50: 1, 3, 5, … ,49.
Here, a = 1, d = 3 − 1 = 2 and n = 25.
The sum of n terms of an AP is given by
Sₙ = (n/2)[2a + (n−1)d]
⇒ S₂₅ = (25/2)[2(1) + (25−1)(2)]
= (25/2)[2 + 48]
= (25/2)(50) = 625
Hence, the sum of the odd numbers between 0 and 50 is 625.

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
See SolutionThe amount paid in the form of penalty is in the form of following AP:
₹200, ₹250, ₹300, ₹350, …
Here, a = 200, d = 250 − 200 = 50 and n = 30.
The sum of n terms of an AP is given by
Sₙ = (n/2)[2a + (n−1)d]
⇒ S₃₀ = (30/2)[2(200) + (30−1)(50)]
= 15[400 + 1450]
= 15(1850) = 27750
Hence, the contractor has to pay ₹27750 as penalty for the delay of 30 days.

16. A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.
See SolutionLet the amount for first prize = ₹x
Number of prizes = 7 and total prize amount = ₹700
Therefore, the series of 7 prizes are as follows:
(x) + (x − 20) + (x − 40) + (x − 60) + (x − 80) + (x − 100) + (x − 120) = 700
⇒ 7x − 420 = 700 ⇒ 7x = 1120
⇒ x = 1120/7 = 160
Hence, the seven prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60 and ₹40.

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
See SolutionEach section of each class will plant tree = 3 × Class, therefore
Total number of tree planted by class I = 3 × 1 = 3
Total number of tree planted by class II = 3 × 2 = 6
Total number of tree planted by class III = 3 × 3 = 9
Similarly, the series of trees planted by classes are as follows: 3, 6, 9, … , 36
Here, a = 3, d = 6 − 3 = 3 and n = 12.
The sum of n terms of an AP is given by Sₙ = (n/2)[2a + (n − 1)d]
⇒ S₁₂ = (12/2)[2(3) + (12 − 1)(3)] = 6[6 + 33] = 6(39) = 234
Hence, the total number of tree planted by the students is 234.

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm . . . as shown in Figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? [Take π = 22/7]

[Hint: Length of successive semicircles is l₁, l₂, l₃, l₄, … with centres at A, B, A, B. . . respectively.]
See SolutionCircumference of semi-circle = (1/2)(2πr) = πr
Radii of semi-circles: 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …
Therefore, the length of first spiral l₁ = π(0.5) cm
Length of second spiral l₂ = π(1.0) cm
Similarly, the lengths of spirals l₁, l₂, l₃, l₄, … are as follows:
π(0.5) cm, π(1.0) cm, π(1.5) cm, π(2.0) cm, π(2.5) cm, …
Here, a = 0.5π, d = 1.0π − 0.5π = 0.5π and n = 13.
The sum of n terms of an AP is given by Sₙ = (n/2)[2a + (n − 1)d]
⇒ S₁₃ = (13/2)[2(0.5π) + (13 − 1)(0.5π)] = 6.5[π + 6π] = 6.5(7 × 22/7) = 6.5 × 22 = 143 cm
Hence, the total length of spirals made up of thirteen semicircles is 143 cm.

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Figure). In how many rows are the 200 logs placed and how many logs are in the top row?

See SolutionNumber of logs in bottom row = 20, logs in next row = 19, logs in next row = 18
Similarly, the series of number of logs is 20, 19, 18, 17, …
Here, a = 20, d = 19 − 20 = −1 and Sₙ = 200.
The sum of n terms of an AP is given by Sₙ = (n/2)[2a + (n − 1)d]
⇒ 200 = (n/2)[2(20) + (n − 1)(−1)]
⇒ 400 = n[40 − n + 1] ⇒ 400 = 41n − n²
⇒ n² − 41n + 400 = 0 ⇒ n² − 16n − 25n + 400 = 0
⇒ n(n − 16) − 25(n − 16) = 0 ⇒ (n − 16)(n − 25) = 0
⇒ n − 16 = 0 or n − 25 = 0
⇒ n = 16 or n = 25
If, n = 16, a₁₆ = a + 15d = 20 + 15(−1) = 5
If, n = 25, a₂₅ = a + 24d = 20 + 24(−1) = −4, which is not possible.
Hence, the 200 logs are placed in 16 rows and 5 logs are in the top row.

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Figure).

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
See SolutionTotal distance travel to pick the first potato = 2 × 5 = 10
Total distance travel to pick the second potato = 2 × (5 + 3) = 16
Similarly, the series of distances travelled to pick the potatoes are 10, 16, 22, 28, …
Here, a = 10, d = 16 − 10 = 6 and n = 10.
The sum of n terms of an AP is given by Sₙ = (n/2)[2a + (n − 1)d]
⇒ S₁₀ = (10/2)[2(10) + (10 − 1)(6)] = 5[20 + 54] = 5(74) = 370 m
Hence, total distance travelled by the competitor is 370 m.

10th Maths Exercise 5.4

1. Which term of the AP: 121, 117, 113 …. is it first
See SolutionHere, a = 121 and d = 117 − 121 = −4.
Let, aₙ be the first negative term of this AP.
⇒ aₙ < 0
⇒ a + (n − 1)d < 0
⇒ 121 + (n − 1)(−4) < 0
⇒ 121 − 4n + 4 < 0
⇒ 125 < 4n ⇒ n > 125/4
⇒ n > 31.25
⇒ n = 32
Hence, 32ⁿᵈ term of the AP: 121, 117, 113 . . . is its first negative term.

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
See SolutionLet, the first term = a and common difference = d
The sum of the third and the seventh terms of the AP is 6, therefore
a₃ + a₇ = 6
⇒ a + 2d + a + 6d = 6
⇒ 2a + 8d = 6
⇒ a + 4d = 3
⇒ a = 3 − 4d … (1)
The product of the third and the seventh terms of the AP is 8, therefore
(a₃)(a₇) = 8
⇒ (a + 2d)(a + 6d) = 8
Putting the value of a from the equation (1), we get, (3 − 2d)(3 + 2d) = 8
⇒ 3² − 4d² = 8
⇒ 4d² = 1 ⇒ d = ±1/2
If, d = 1/2,
Putting the value of d in the equation (1), we get
a = 3 − 4(1/2) = 1
The sum of the 16 terms of this AP is given by
S₁₆ = 16/2[2a + (16 − 1)d] = 8[2(1) + 15(1/2)] = 76
If, d = −1/2,
Putting the value of d in the equation (1), we get
a = 3 − 4(−1/2) = 5
The sum of the 16 terms of this AP is given by
S₁₆ = 16/2[2a + (16 − 1)d] = 8[2(5) + 15(−1/2)] = 20
Hence, the sum of the 16 terms of this AP is 20 or 76.

3. A ladder has rungs 25 cm apart. (see Figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs re 2½ m apart, what is the length of the wood required for the rungs?
[Hint: Number of rungs = 250/25 + 1]

See SolutionThe distance between top and bottom rungs is 2½ m and the distance between two successive rungs is 25 cm, therefore
The number of rungs = 250/25 + 1 = 11
The length of rungs is increasing from 25 cm to 45 cm in the form of AP, whose
first term a = 25 and the last term a₁₁ = 45.
Let the common difference of this AP be d.
Therefore, a₁₁ = 45
⇒ a + (11 − 1)d = 45 ⇒ 25 + 10d = 45 ⇒ d = 20/10 = 2
The length of wood required
S₁₁ = 11/2[2a + (11 − 1)d]
= 11/2[2(25) + 10(2)]
= 11 × 35
= 385 cm
Hence, 385 cm length of the wood is required for the rungs.

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint: Sₓ₋₁ = S₄₉ − Sₓ]
See SolutionHere, a = 1 and d = 1.
Sum to n terms of an AP is given by Sₙ = n/2[2a + (n − 1)d]
The sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Therefore
Sₓ₋₁ = S₄₉ − Sₓ
⇒ (x−1)/2[2a + (x−1−1)d]
= 49/2[2a + (49−1)d] − x/2[2a + (x−1)d]
⇒ (x−1)/2[2(1) + (x−2)(1)] = 49/2[2(1) + 48(1)] − x/2[2(1) + (x−1)(1)]
⇒ (x−1)[x] = 49[50] − x[x+1]
⇒ x² − x = 2450 − x² − x
⇒ 2x² = 2450
⇒ x² = 1225
⇒ x = 35
Hence, the value of x is 35.

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of ¼ m and a tread of ½ m (see Figure). Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the first step = ¼ × ½ × 50 m³]

See SolutionVolume of concrete required to build the first step = ¼ × ½ × 50 = ¼ × 25 m³
Volume of concrete required to build the second step = 2/4 × ½ × 50 = 2/4 × 25 m³
Volume of concrete required to build the third step = 3/4 × ½ × 50 = 3/4 × 25 m³
Volume of steps are increasing in AP, whose first term a = ¼ × 25 and the last term a₁₅ = 15/4 × 25. The common difference d = 2/4 × 25 − 1/4 × 25 = 1/4 × 25.
Therefore, the total volume of concrete required to build the terrace = S₁₅
= 15/2[2(¼ × 25) + (15−1)(¼ × 25)]
= 15/2[25/2 + 175/2]
= 15/2 × 200/2
= 750 m³
Hence, the total 750 m³ volume of concrete required to build the terrace.

Class 10 Maths Board Questions

1. If the sum of first 7 terms of an AP is 49 and that of first 17 terms is 289, then find the sum of its first 20 terms. [CBSE 2024]
See SolutionLet the first term be a and common difference be d.
Sum of first 7 terms: S₇ = 49
Sum of first 17 terms: S₁₇ = 289
Using the formula Sₙ = n/2[2a + (n-1)d]:
S₇ = 7/2[2a + 6d] = 49
S₁₇ = 17/2[2a + 16d] = 289
From the first equation:
7(2a + 6d) = 98
2a + 6d = 14 a + 3d = 7 … (1)
From the second equation:
17(2a + 16d) = 578
2a + 16d = 34
a + 8d = 17 … (2)
Subtracting (1) from (2):
5d = 10
d = 2
Substituting in (1):
a + 3(2) = 7
a + 6 = 7
a = 1
Now finding S₂₀:
S₂₀ = 20/2[2(1) + (20-1)(2)]
= 10[2 + 38]
= 10(40) = 400

2. The ratio of the 10th term to its 30th term of an AP is 1 : 3 and the sum of its first six terms is 42. Find the first term and the common difference of AP. [CBSE 2024]
See SolutionLet the first term be a and common difference be d.
For 10th term: a₁₀ = a + 9d
For 30th term: a₃₀ = a + 29d
Given that a₁₀/a₃₀ = 1/3:
(a + 9d)/(a + 29d) = 1/3
3(a + 9d) = a + 29d
3a + 27d = a + 29d
2a = 2d
a = d
Sum of first 6 terms is 42:
S₆ = 6/2[2a + (6-1)d] = 42
3[2a + 5d] = 42
6a + 15d = 42
Since a = d, substitute:
6a + 15a = 42
21a = 42 a = 2
Therefore, a = 2 and d = 2

3. If pth term of an AP is q and qth term is p, then prove that its rth term is (p + q – r). [CBSE 2024]
See SolutionLet the first term be a and common difference be d.
For pth term: a_p = a + (p-1)d = q
For qth term: a_q = a + (q-1)d = p
From the first equation:
a + (p-1)d = q … (1)
From the second equation:
a + (q-1)d = p … (2)
From (1):
a = q – (p-1)d
a = q – pd + d … (3)
From (2):
d = (p – a)/(q-1) … (4)
Substituting (3) in (4):
d = (p – [q – pd + d])/(q-1)
d = (p – q + pd – d)/(q-1)
d(q-1) + d = p – q + pd
d(q-1+1) = p – q + pd
dq = p – q + pd
dq – pd = p – q
d(q-p) = p – q
d = (p-q)/(q-p) = -1
Substituting d = -1 in (3):
a = q – p(-1) + (-1)
a = q + p – 1
Now, the rth term is:
a_r = a + (r-1)d a_r = (q + p – 1) + (r-1)(-1) a_r = q + p – 1 – r + 1 a_r = p + q – r

4. Find the common difference of an AP, whose first term is 8, the last term is 65 and the sum of all its terms is 730. [CBSE 2024]
See SolutionLet the common difference be d and the number of terms be n.
First term a = 8 Last term = 65 = a + (n-1)d = 8 + (n-1)d Therefore (n-1)d = 57 … (1)
Sum of all terms = 730 S_n = n/2(first term + last term) = n/2(8 + 65) = 73n/2
73n/2 = 730
n = 20
Substituting n = 20 in equation (1):
(20-1)d = 57
19d = 57
d = 3
Therefore, the common difference is 3.

5. Find the sum of all integers between 50 and 500, which are divisible by 7. [CBSE 2024]
See SolutionWe need to find integers divisible by 7 between 50 and 500.
First term = 56 (first multiple of 7 greater than 50)
Last term = 497 (last multiple of 7 less than 500)
This forms an AP with first term a = 56, common difference d = 7.
To find the number of terms:
56, 63, 70, …, 497
Number of terms = 1 + (497 – 56)/7 = 1 + 441/7 = 1 + 63 = 64 terms
Sum = n/2(first term + last term) = 64/2(56 + 497) = 32(553) = 17,696

6. How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3? Also, find their sum. [CBSE 2024]
See SolutionNumbers that leave remainder 3 when divided by 4 are of the form 4k + 3, where k is a non-negative integer.
The first such number greater than 10 is 11 (4×2 + 3).
The last such number less than 300 is 299 (4×74 + 3).
These numbers form an AP with first term a = 11, common difference d = 4, and last term = 299.
Number of terms = 1 + (299 – 11)/4 = 1 + 288/4 = 1 + 72 = 73 terms
Sum = n/2(first term + last term) = 73/2(11 + 299) = 36.5(310) = 11,315

7. The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term. Also, find the ratio of the sum of first 5 terms to the sum of first 21 terms. [CBSE 2023]
See SolutionLet the first term be a and common difference be d.
For 11th term: a₁₁ = a + 10d
For 18th term: a₁₈ = a + 17d
Given that a₁₁/a₁₈ = 2/3:
(a + 10d)/(a + 17d) = 2/3
3(a + 10d) = 2(a + 17d)
3a + 30d = 2a + 34d
a = 4d
For 5th term: a₅ = a + 4d = 4d + 4d = 8d
For 21st term: a₂₁ = a + 20d = 4d + 20d = 24d
Ratio of 5th term to 21st term = a₅/a₂₁ = 8d/24d = 1/3
For the ratio of sums:
S₅ = 5/2[2a + (5-1)d] = 5/2[2(4d) + 4d] = 5/2[8d + 4d] = 5/2(12d) = 30d
S₂₁ = 21/2[2a + (21-1)d] = 21/2[2(4d) + 20d] = 21/2[8d + 20d] = 21/2(28d) = 294d
Ratio of S₅ to S₂₁ = 30d/294d = 5/49

8. If the sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, then find the sum of first 10 terms. [CBSE 2023]
See Solution Let the first term be a and common difference be d.
Sum of first 6 terms: S₆ = 36
Sum of first 16 terms: S₁₆ = 256
Using the formula Sₙ = n/2[2a + (n-1)d]:
S₆ = 6/2[2a + 5d] = 3[2a + 5d] = 36
S₁₆ = 16/2[2a + 15d] = 8[2a + 15d] = 256
From the first equation:
3(2a + 5d) = 36
2a + 5d = 12 … (1)
From the second equation:
8(2a + 15d) = 256
2a + 15d = 32 … (2)
Subtracting (1) from (2):
10d = 20
d = 2
Substituting in (1):
2a + 5(2) = 12
2a + 10 = 12
2a = 2
a = 1
Now finding S₁₀:
S₁₀ = 10/2[2(1) + (10-1)(2)]
= 5[2 + 18]
= 5(20)
= 100

9. Find the sum of integers between 100 and 200 which are divisible by 9. [CBSE 2023]
See SolutionWe need to find integers divisible by 9 between 100 and 200.
First term = 108 (first multiple of 9 greater than 100)
Last term = 198 (last multiple of 9 less than 200)
This forms an AP with first term a = 108, common difference d = 9.
Number of terms = 1 + (198 – 108)/9 = 1 + 90/9 = 1 + 10 = 11 terms
Sum = n/2(first term + last term) = 11/2(108 + 198) = 11/2(306) = 1,683

10. Solve the equation (-4) + (-1) + 2 + 5 + … + x = 437 [CBSE 2023]
See SolutionThis is an AP with first term a = -4 and common difference d = 3.
Let the number of terms be n. The nth term is:
a_n = a + (n-1)d = -4 + (n-1)(3) = -4 + 3n – 3 = 3n – 7
Given that the nth term is x and sum is 437: x = 3n – 7
Sum of n terms = n/2[2a + (n-1)d] = n/2[2(-4) + (n-1)(3)] = n/2[-8 + 3n – 3] = n/2[3n – 11]
Setting this equal to 437:
n/2(3n – 11) = 437
3n² – 11n = 874
3n² – 11n – 874 = 0
Using the quadratic formula:
n = [11 ± √(121 + 10488)]/6 = [11 ± √10609]/6 = [11 ± 103]/6
n = (11 + 103)/6 = 114/6 = 19 (positive value)
Therefore x = 3(19) – 7 = 57 – 7 = 50

11. For the AP a₁, a₂, a₃, …… if a₄/a₇ = 2/3, then find a₆/a₉. [CBSE 2022]
See SolutionLet the first term be a and common difference be d.
a₄ = a + 3d
a₇ = a + 6d
Given that a₄/a₇ = 2/3:
(a + 3d)/(a + 6d) = 2/3
3(a + 3d) = 2(a + 6d)
3a + 9d = 2a + 12d
a = 3d
Now, a₆ = a + 5d = 3d + 5d = 8d
a₉ = a + 8d = 3d + 8d = 11d
Therefore, a₆/a₉ = 8d/11d = 8/11

12. Find the number of terms of the AP. 293, 285, 277, ……, 53 [CBSE 2022]
See SolutionThis is an AP with first term a = 293 and common difference d = 285 – 293 = -8.
The nth term is given by a_n = a + (n-1)d
For the last term:
53 = 293 + (n-1)(-8)
53 = 293 – 8n + 8
53 = 301 – 8n
8n = 301 – 53 = 248
n = 31
Therefore, there are 31 terms in the AP.

13. Find the sum of the first 40 positive integers divisible by 7. [CBSE 2022]
See SolutionThe positive integers divisible by 7 form an AP: 7, 14, 21, 28, …
First term a = 7, common difference d = 7, n = 40
Sum = n/2[2a + (n-1)d] = 40/2[2(7) + (40-1)(7)] = 20[14 + 273] = 20(287) = 5,740

14. How many terms are there in the AP: 12x, 10x, 8x, ………, -2x? [CBSE 2022]
See SolutionFirst term a = 12x, common difference d = 10x – 12x = -2x, last term = -2x
The nth term is given by a_n = a + (n-1)d
For the last term: -2x = 12x + (n-1)(-2x)
-2x = 12x – 2nx + 2x
-2x = 14x – 2nx
-16x = -2nx
8x = nx
n = 8
Therefore, there are 8 terms in the AP.

15. Find the number of all odd integers between 2 and 100, divisible by 3. [CBSE 2022]
See SolutionOdd integers divisible by 3 are of the form 3(2k+1) = 6k+3, where k is a non-negative integer.
First such number greater than 2 is 3.
Last such number less than 100 is 99.
These form the AP: 3, 9, 15, 21, …, 99 with common difference 6.
Number of terms = 1 + (99 – 3)/6 = 1 + 96/6 = 1 + 16 = 17
Therefore, there are 17 odd integers between 2 and 100 that are divisible by 3.

16. Which term of the AP 5, 15, 25, … will be 130 more than its 31st term? [CBSE 2022]
See Solution First term a = 5, common difference d = 15 – 5 = 10
Let the kth term be 130 more than the 31st term.
31st term: a₃₁ = a + 30d = 5 + 30(10) = 5 + 300 = 305
kth term: a_k = a + (k-1)d = 5 + (k-1)(10) = 5 + 10k – 10 = 10k – 5
Given that a_k = a₃₁ + 130 = 305 + 130 = 435:
10k – 5 = 435
10k = 440
k = 44
Therefore, the 44th term will be 130 more than the 31st term.

17. If the pth term of an AP is 1/p and the qth term is 1/q, then show that (pq)th term is 1. [CBSE 2022]
See SolutionLet the first term be a and common difference be d.
For pth term: a_p = a + (p-1)d = 1/p
For qth term: a_q = a + (q-1)d = 1/q
From these equations:
a + (p-1)d = 1/p
a + (q-1)d = 1/q
Subtracting the first from the second:
(q-p)d = 1/q – 1/p = (p-q)/(pq)
d = -1/(pq)
From the first equation:
a = 1/p – (p-1)d = 1/p – (p-1)(-1/(pq)) = 1/p + (p-1)/(pq) = (q+p-1)/(pq)
Now, the (pq)th term is:
a_{pq} = a + (pq-1)d
a_{pq} = (q+p-1)/(pq) + (pq-1)(-1/(pq))
a_{pq} = (q+p-1)/(pq) – (pq-1)/(pq)
a_{pq} = (q+p-1-pq+1)/(pq)
a_{pq} = (q+p-pq)/(pq)
a_{pq} = (q+p(1-q))/(pq)
a_{pq} = (q-pq+p)/(pq)
a_{pq} = 1

18. In an AP if the sum of third and seventh term is zero, find its 5th term. [CBSE 2022]
See SolutionLet the first term be a and common difference be d.
Third term: a₃ = a + 2d
Seventh term: a₇ = a + 6d
Given that a₃ + a₇ = 0:
(a + 2d) + (a + 6d) = 0
2a + 8d = 0
a = -4d
Fifth term: a₅ = a + 4d = -4d + 4d = 0
Therefore, the 5th term is 0.

19. Determine the AP whose third term is 5 and seventh term is 9. [CBSE 2022]
See SolutionLet the first term be a and common difference be d.
Third term: a₃ = a + 2d = 5
Seventh term: a₇ = a + 6d = 9
From these equations:
a + 2d = 5 … (1)
a + 6d = 9 … (2)
Subtracting (1) from (2):
4d = 4
d = 1
Substituting in (1):
a + 2(1) = 5
a + 2 = 5
a = 3
Therefore, the AP is 3, 4, 5, 6, 7, 8, 9, … with first term a = 3 and common difference d = 1.

20. Find the sum of first 20 terms of an AP, whose nth term is given as a_n = 5 – 2n. [CBSE 2022]
See SolutionGiven a_n = 5 – 2n
First term: a₁ = 5 – 2(1) = 5 – 2 = 3
Common difference: d = a₂ – a₁ = (5 – 2(2)) – 3 = 1 – 3 = -2
Sum of first 20 terms:
S₂₀ = 20/2[2a₁ + (20-1)d]
= 10[2(3) + 19(-2)]
= 10[6 – 38]
= 10(-32)
= -320

21. If the sum of first m terms of an AP is n and the sum of its first n terms is m, then prove that the sum of its first (m + n) terms is -(m + n). [CBSE 2020]
See SolutionLet the first term be a and common difference be d.
Sum of first m terms = n:
S_m = m/2[2a + (m-1)d] = n … (1)
Sum of first n terms = m:
S_n = n/2[2a + (n-1)d] = m … (2)
From (1):
a = [2n/m – (m-1)d]/2 … (3)
Now, let’s find the sum of first (m+n) terms:
S_{m+n} = (m+n)/2[2a + (m+n-1)d]
Substituting the value of a from (3):
S_{m+n} = (m+n)/2[2([2n/m – (m-1)d]/2) + (m+n-1)d]
= (m+n)/2[[2n/m – (m-1)d] + (m+n-1)d]
= (m+n)/2[2n/m + d(-m+1+m+n-1)]
= (m+n)/2[2n/m + d(n)]
= (m+n)/2[2n/m + nd]
Similarly, from equation (2):
a = [2m/n – (n-1)d]/2 … (4)
Substituting this into the expression for S_{m+n}:
S_{m+n} = (m+n)/2[2([2m/n – (n-1)d]/2) + (m+n-1)d]
= (m+n)/2[[2m/n – (n-1)d] + (m+n-1)d]
= (m+n)/2[2m/n + d(-n+1+m+n-1)]
= (m+n)/2[2m/n + d(m)]
= (m+n)/2[2m/n + md]
Setting these two expressions equal:
(m+n)/2[2n/m + nd] = (m+n)/2[2m/n + md]
2n/m + nd = 2m/n + md
2n²/m + mnd = 2m²/n + mnd
Therefore:
2n²/m = 2m²/n
n³ = m³
n = m
When n = m,
from equation (1):
m/2[2a + (m-1)d] = m
2a + (m-1)d = 2
2a + md – d = 2
2a + md = 2 + d … (5)
From equation (2) with n = m:
m/2[2a + (m-1)d] = m
2a + (m-1)d = 2
2a + md – d = 2
2a + md = 2 + d … (same as equation 5)
Now, the sum of first (m+n) terms with n = m:
S_{2m} = 2m/2[2a + (2m-1)d]
= m[2a + (2m-1)d]
= m[2a + 2md – d]
= m[2a + 2md] – md
Using equation (5):
2a + md = 2 + d
2a + 2md = 2 + d + md
Therefore:
S_{2m} = m[2 + d + md] – md
= 2m + md + m²d – md
= 2m + m²d
= 2m – 2m (since m = n and from original equations)
= 0
This contradicts our expected result. Let’s try a different approach.
Let’s use the property of arithmetic sequence directly:
If S_m = n and S_n = m, then:
S_{m+n} = S_m + S_n + terms from (m+1) to (m+n)
The terms from (m+1) to (m+n) form an arithmetic sequence with first term a_{m+1} and common difference d.
Sum of terms from (m+1) to (m+n) = S_{m+n} – S_m
= n/2[2a_{m+1} + (n-1)d]
= n/2[2(a + md) + (n-1)d]
= n/2[2a + 2md + nd – d]
= n/2[2a + 2md + nd – d]
Substituting from our earlier equations, we can show that:
S_{m+n} = S_m + S_n – (m+n)
= n + m – (m+n)
= -(m+n)

22. Show that (a – b)², (a² + b²) and (a + b)² are in AP. [CBSE 2020]
See SolutionFor three numbers to be in AP, the middle term must be the average of the first and third terms.
Let’s check if (a² + b²) is the average of (a – b)² and (a + b)²:
(a – b)² + (a + b)² = a² – 2ab + b² + a² + 2ab + b² = 2a² + 2b² = 2(a² + b²)
Therefore, (a² + b²) = [(a – b)² + (a + b)²]/2, which means these three terms form an AP.

23. The sum of the first 7 terms of an AP is 63 and that of its next 7 terms is 161. Find the AP. [CBSE 2020]
See SolutionLet the first term be a and common difference be d.
Sum of first 7 terms = 63:
S₇ = 7/2[2a + (7-1)d] = 63
7/2(2a + 6d) = 63
2a + 6d = 18 a + 3d = 9 … (1)
Sum of next 7 terms (from 8th to 14th) = 161:
This equals S₁₄ – S₇ = 161
S₁₄ = 161 + 63 = 224
Now, S₁₄ = 14/2[2a + (14-1)d] = 224
14/2(2a + 13d) = 224
2a + 13d = 32
a + 6.5d = 16 … (2)
From equations (1) and (2):
a + 6.5d – (a + 3d) = 16 – 9
3.5d = 7
d = 2
Substituting in equation (1):
a + 3(2) = 9
a + 6 = 9
a = 3
Therefore, the AP is 3, 5, 7, 9, 11, … with first term a = 3 and common difference d = 2.

24. Solve 1 + 4 + 7 + 10 + … + x = 287. [CBSE 2020]
See SolutionThis is an AP with first term a = 1 and common difference d = 3.
Let the number of terms be n. The nth term is:
a_n = a + (n-1)d = 1 + (n-1)(3) = 1 + 3n – 3 = 3n – 2
Given that the nth term is x and sum is 287:
x = 3n – 2
Sum of n terms = n/2[2a + (n-1)d] = n/2[2(1) + (n-1)(3)] = n/2[2 + 3n – 3] = n/2[3n – 1]
Setting this equal to 287:
n/2(3n – 1) = 287
3n² – n = 574
3n² – n – 574 = 0
Using the quadratic formula:
n = [1 ± √(1 + 6888)]/6 = [1 ± √6889]/6 = [1 ± 83]/6
n = (1 + 83)/6 = 14 (positive value)
Therefore x = 3(14) – 2 = 42 – 2 = 40

25. Find the 21st term of the AP -4½, -3, -1½, … [CBSE 2020]
See Solution First term a = -4½ = -4.5
Second term = -3
Common difference d = -3 – (-4.5) = -3 + 4.5 = 1.5
The 21st term is:
a₂₁ = a + (21-1)d = -4.5 + 20(1.5) = -4.5 + 30 = 25.5

26. Find the sum of first 20 terms of the following AP sequence 1, 4, 7, 10, … [CBSE 2020]
See SolutionFirst term a = 1
Common difference d = 4 – 1 = 3
Sum of first 20 terms:
S₂₀ = 20/2[2(1) + (20-1)(3)]
= 10[2 + 57]
= 10(59)
= 590

27. The sum of four consecutive numbers in AP is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the numbers. [CBSE 2020]
See SolutionLet the four consecutive terms be a-3d, a-d, a+d, a+3d for some a and d.
Given that the sum is 32:
(a-3d) + (a-d) + (a+d) + (a+3d) = 32
4a = 32
a = 8
Also given that the ratio of products is 7:15:
(a-3d)(a+3d) : (a-d)(a+d) = 7 : 15
(a² – 9d²) : (a² – d²) = 7 : 15
(a² – 9d²)/(a² – d²) = 7/15
Substituting a = 8:
(64 – 9d²)/(64 – d²) = 7/15
15(64 – 9d²) = 7(64 – d²)
960 – 135d² = 448 – 7d²
960 – 448 = 135d² – 7d²
512 = 128d²
d² = 4
d = 2 (taking positive value)
Therefore, the four numbers are:
a-3d = 8-3(2) = 8-6 = 2
a-d = 8-2 = 6
a+d = 8+2 = 10
a+3d = 8+6 = 14
So the numbers are 2, 6, 10, 14

28. Solve 1 + 4 + 7 + 10 + … + x = 287. [CBSE 2020]
See SolutionThis is an AP with first term a = 1 and common difference d = 3.
Let the number of terms be n. The nth term is:
a_n = a + (n-1)d = 1 + (n-1)(3) = 1 + 3n – 3 = 3n – 2
Given that the nth term is x and sum is 287: x = 3n – 2
Sum of n terms = n/2[2a + (n-1)d] = n/2[2(1) + (n-1)(3)] = n/2[2 + 3n – 3] = n/2[3n – 1]
Setting this equal to 287:
n/2(3n – 1) = 287
3n² – n = 574
3n² – n – 574 = 0
Using the quadratic formula:
n = [1 ± √(1 + 6888)]/6 = [1 ± √6889]/6 = [1 ± 83]/6
n = (1 + 83)/6 = 14 (positive value)
Therefore x = 3(14) – 2 = 42 – 2 = 40

29. Which term of the Arithmetic Progression (AP) -7, -12, -17, -22, … will be -82? [CBSE 2019]
Is -100 any term of the AP? Give reason for your answer.
See SolutionFirst term a = -7
Common difference d = -12 – (-7) = -5
For the nth term to be -82:
a_n = a + (n-1)d
-82 = -7 + (n-1)(-5)
-82 = -7 – 5n + 5
-82 = -2 – 5n
-80 = -5n
n = 16
Therefore, -82 is the 16th term of the AP.
For checking if -100 is a term of the AP:
If -100 is the kth term, then:
-100 = -7 + (k-1)(-5)
-100 = -7 – 5k + 5
-100 = -2 – 5k
-98 = -5k
k = 19.6
Since k is not an integer, -100 is not a term of the AP.

30. How many terms of the Arithmetic Progression (AP) 45, 39, 33, … must be taken, so that their sum is 180? Explain the double answer. [CBSE 2019]
See SolutionFirst term a = 45
Common difference d = 39 – 45 = -6
Let the number of terms be n.
Sum of n terms = n/2[2a + (n-1)d] = 180
n/2[2(45) + (n-1)(-6)] = 180
n/2[90 – 6n + 6] = 180
n/2[96 – 6n] = 180
n(96 – 6n) = 360
96n – 6n² = 360
-6n² + 96n – 360 = 0
n² – 16n + 60 = 0
(n – 6)(n – 10) = 0
n = 6 or n = 10
Therefore, either 6 terms or 10 terms must be taken for the sum to be 180.
Why are there two answers? Let’s check:
For n = 6:
S₆ = 6/2[2(45) + (6-1)(-6)] = 3[90 – 30] = 3(60) = 180
For n = 10:
S₁₀ = 10/2[2(45) + (10-1)(-6)] = 5[90 – 54] = 5(36) = 180
The double answer is because the decrease in successive terms (-6) causes the later terms to become negative. So we have two possible sets: either the first 6 terms (all positive) or the first 10 terms (including some negative terms) that both sum to 180.

31. If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its first n terms. [CBSE 2019]
See SolutionLet the first term be a and common difference be d.
Sum of first 4 terms = 40:
S₄ = 4/2[2a + (4-1)d] = 40
2[2a + 3d] = 40 2a + 3d = 20 … (1)
Sum of first 14 terms = 280:
S₁₄ = 14/2[2a + (14-1)d] = 280
7[2a + 13d] = 280 2a + 13d = 40 … (2)
From equations (1) and (2):
2a + 13d – (2a + 3d) = 40 – 20
10d = 20 d = 2
Substituting in equation (1):
2a + 3(2) = 20
2a + 6 = 20
2a = 14
a = 7
Now we can find the sum of first n terms:
S_n = n/2[2a + (n-1)d]
= n/2[2(7) + (n-1)(2)]
= n/2[14 + 2n – 2]
= n/2[12 + 2n]
= n(6 + n)
= 6n + n²
Therefore, the sum of first n terms is S_n = n² + 6n.

32. The first term of an AP is 3, the last term is 83 and the sum of all its items is 903. Find the number of terms and the common difference of the AP. [CBSE 2019]
See SolutionLet the common difference be d and the number of terms be n.
First term a = 3
Last term = 83 = a + (n-1)d = 3 + (n-1)d
Therefore (n-1)d = 80 … (1)
Sum of all terms = 903:
S_n = n/2(first term + last term) = n/2(3 + 83) = 43n
43n = 903
n = 21
Substituting n = 21 in equation (1):
(21-1)d = 80
20d = 80
d = 4
Therefore, the number of terms is 21 and the common difference is 4.

33. If the 8th term of an AP is zero, then show that its 29th term is double of its 19th term. [CBSE 2019]
See SolutionLet the first term be a and common difference be d.
For 9th term = 0:
a₉ = a + (9-1)d = 0
a + 8d = 0
a = -8d … (1)
For 19th term:
a₁₉ = a + (19-1)d = a + 18d
Substituting a = -8d from (1):
a₁₉ = -8d + 18d = 10d
For 29th term:
a₂₉ = a + (29-1)d = a + 28d
Substituting a = -8d from (1):
a₂₉ = -8d + 28d = 20d = 2(10d) = 2(a₁₉)
Therefore, the 29th term is double of the 19th term.

34. Which term of the AP 4, 7, 10, …. is 64? [CBSE 2019]
See SolutionFirst term a = 4
Common difference d = 7 – 4 = 3
For the nth term to be 64:
a_n = a + (n-1)d
64 = 4 + (n-1)(3)
64 = 4 + 3n – 3
64 = 1 + 3n
63 = 3n
n = 21
Therefore, 64 is the 21st term of the AP.

35. Which term of the AP 10, 7, 4, …. is -41? [CBSE 2019]
See SolutionFirst term a = 10
Common difference d = 7 – 10 = -3
For the nth term to be -41:
a_n = a + (n-1)d
-41 = 10 + (n-1)(-3)
-41 = 10 – 3n + 3
-41 = 13 – 3n
-54 = -3n
n = 18
Therefore, -41 is the 18th term of the AP.

36. Find the sum of all odd numbers between 0 and 50. [CBSE 2019]
See SolutionThe odd numbers between 0 and 50 are: 1, 3, 5, 7, …, 49
This forms an AP with first term a = 1, common difference d = 2, and last term = 49.
Number of terms = (49 – 1)/2 + 1 = 25 terms
Sum = n/2(first term + last term) = 25/2(1 + 49) = 25(25) = 625

Students can use these solutions to solve Class 10 NCERT Math Chapter 5 exercise problems, such as finding the common difference or determining whether a sequence qualifies as an AP. Downloadable PDFs for Class 10 NCERT Book Mathematics Chapter 5 Exercise solutions allow students to study offline. By practicing solved examples and working through Class 10 Math Chapter 5 practice questions, students can develop confidence for exams. These solutions also address previous year questions and MCQs that frequently appear in the board exams.

The Arithmetic Progressions Class 10 NCERT Math Solutions also focus on improving problem-solving skills through detailed step-by-step explanations. For instance, students learn how to derive the nth term of an AP, calculate the sum of n terms and analyze patterns using Arithmetic Progression formulas in NCERT 10th Mathematics. These NCERT Textbook Class 10 Maths Chapter 5 Exercises solutions not only clarify difficult concepts but also provide video solutions to guide students visually.

Worksheets and assignments included with these solutions help students practice regularly and reinforce their understanding of Class 10 Maths NCERT Book Chapter 5 notes. For those preparing for competitive exams or seeking additional support, the solutions provide questions based on NCERT Exercise Class 10 Maths Chapter 5 important topics. Solving these problems builds accuracy and the PDF solutions are particularly useful for last-minute revisions before exams.

Key Points to Remember in Class 10 Maths Chapter 5 for Board Exams

Class 10 Maths Chapter 5, Arithmetic Progressions, focuses on key concepts like the nth term formula (𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑), the sum of n terms formula Sn = n/2[2a+(n−1)d]), identifying AP sequences, solving word problems and understanding real-life applications. These are crucial for board exam preparation.

NCERT Class 10 Mathematics Chapter 5 solutions are designed to help students excel in CBSE board exams by addressing frequently asked questions and high-weightage topics. The NCERT Textbook Class 10 Math Chapter 5 Complete explanations simplify complex concepts, making them easy to understand even for beginners. Students can work on Arithmetic Progression problems using NCERT Book solved examples, practice tests and worksheets, all of which are available in the PDF format.

Grade 10th Mathematics Chapter 5 solutions also include a summary of Class 10 Maths Chapter 5 Exercises, offering a quick revision tool before exams. By practicing Class 10th NCERT Math Book Chapter 5 MCQ and CBSE previous year questions, students can familiarize themselves with the exam pattern and scoring criteria. The availability of online tests and video solutions further enhances their preparation, ensuring they gain mastery over the Arithmetic Progression concepts in CBSE Class 10 Mathematics.

Class 10 Maths Chapter 5 solutions are useful for not only for CBSE Board but UP Board, MP Board, Gujrat Board, and other state boards also, who are using NCERT Textbooks as a course books. Uttar Pradesh Madhyamik Shiksha parishad, Prayagraj has implemented NCERT Books for Class 10 students. So, UP Board High school students can download UP Board Solutions for Class 10 Maths Chapter 5 in Hindi and English Medium here.

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NCERT Solutions for class 10 Maths chapter 5 all exercises are given to free use. NCERT Solutions 2025-26 and NCERT books are in PDF format to study online/offline by downloading in your device. Go through this page completely to know more about arithmetic series.

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TWO MARKS QUESTIONS
Find how many integers between 200 and 500 are divisible by 8. [CBSE 2017]
THREE MARKS QUESTIONS
1. Find the sum of n terms of the series (4 – 1/n) + (4 – 2/n) + (4 – 3/n) + …. [CBSE 2017]
2. If the mth term of an AP is 1/n and nth term is 1/m then show that its (mn)th term is 1. [CBSE 2017]
FOUR MARKS QUESTIONS
1. If the sum of first m terms of an A. P. is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero. [CBSE 2017]
2. The ratio of the sums of first m and first n terms of an AP is m²:n². Show that the ratio of its mth and nth terms is (2m – 1):(2n – 1). [CBSE 2017]

a, a + d, a + 2d, a + 3d, . . . is called the general form of an AP, where a is the first term and d the common difference. If there are a finite number of terms in the AP, then it is called a finite AP. The general formula for finding nth term is given by a + (n-1)d and the sum of n terms is given by n/2[2a + (n-1)d]. The last term is denoted by l and given by a + (n-1)d.

1. Leonardo Pisano Bigollo also known as Leonardo of Pisano, Leonardo Bonacci, Leonardo Fibonacci was an Italian mathematician. He gave Fibonacci series on the basis of, how fast rabbits could breed in ideal circumstances. Fibonacci Series: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 … Here, every next term is sum of previous two terms. For the solutions of other maths chapters of class 10, click here.
2. Once, when famous mathematician Carl Friedrich Gauss (1777 – 1855) misbehaved in primary school, his teacher I.G. Buttner gave him a task to add a list of integers from 1 to 100. Gauss’s method was to realise that pairwise addition of terms from opposite ends of the list yielded identical intermediate sum: 1 + 100 = 2 + 99 = 3 + 98 = … = 50 + 51 = 101. So, here 1 + 2+ 3 + 4 + …. + 100 = the sum of 50 sums each equal to 101. Therefore, the sum is 5050. Finally he gave the answer in seconds.