NCERT solutions for Class 10 Maths Chapter 5

NCERT solutions for class 10 Maths chapter 5 all exercises of AP – Arithmetic Progression updated for new session 2020-21. These solutions are useful for not only for CBSE Board but UP Board, MP Board, Gujrat Board, etc. 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. All the Online and Offline Apps, NCERT solutions are based on latest NCERT Books 2020-2021.

These apps are applicable for UP Board (Higher Secondary) with CBSE Board and other boards who are using NCERT Books 2020-21 in Hindi and English medium. Download Class 10 Maths Apps based on updated NCERT Solutions for new academic session 2020-21.

NCERT solutions for class 10 Maths chapter 5

Class: 10	Maths (English and Hindi Medium)
Chapter 5:	Arithmetic Progression

10th Maths Chapter 5 Solutions

NCERT Solutions for class 10 Maths chapter 5 all exercises are given below to free use. NCERT Solutions 2020-21 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. Be connected with us to get the latest update regarding to CBSE solutions, Sample Papers, board questions and all other study material related to class 10 Maths solutions of AP.

Class 10 Maths Exercise 5.1 & 5.2 Solutions in Video

Class 10 Maths Exercise 5.1 Solutions in Video

Class 10 Maths Exercise 5.2 Solutions in Video

Class 10 Maths Exercise 5.3 & 5.4 Solutions in Video

Class 10 Maths Exercise 5.3 Solutions in Video

Class 10 Maths Exercise 5.4 Solutions in Video

NCERT Solutions for Class 10 Maths Chapter 5

NCERT Solutions 10th Maths
Exercise 5.1Learn more
NCERT Solutions 10th Maths
Exercise 5.2Learn more
NCERT Solutions 10th Maths
Exercise 5.3Learn more
NCERT Solutions 10th Maths
Exercise 5.4Learn more

What is an Arithmetic Progression (AP)?

An arithmetic progression (AP) is a list (or pattern or series) of numbers in which each next term is obtained by adding or subtracting a fixed number to the preceding term except the first term. This fixed number is called the common difference of the AP, it may be positive, negative or zero.

What are the objective of studying Arithmetic Progression?

Objective of studying Arithmetic Progression – AP
To identify arithmetic progression from a given list of numbers, to determine the general term of an arithmetic progression and to find the sum of first n terms of an arithmetic progression.

Previous Years Questions

TWO MARKS QUESTIONS
1. 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]

About Arithmetic Progression – AP

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.

Historical Facts!

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.
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.

Important Questions on Class 10 Maths Chapter 5

A.P.: 121,117,113,…, का कौन-सा सबसे पहला ऋणात्मक पद होगा?

यहाँ, a = 121 तथा d = 117-121 = -4 है,
माना, A.P. का nवाँ पद पहला ऋणात्मक पद है।
〖⇒a〗_n < 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
अतः, इस A.P. का 32वाँ पद सबसे पहला ऋणात्मक पद होगा।

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Here, a = 5,a_n = 45 and S_n = 400.
The sum of n terms of an AP is given by
S_n = n/2 [a + a_n]
⇒ 400 = n/2 [5 + 45]
⇒ 400 = 25n
⇒ n = 400/25 = 16
a_n = 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.

A.P.: 3,8,13,18,… का कौन सा पद 78 है?

यहाँ, a = 3 तथा d = 8 – 3 = 5 है,
माना, A.P. का nवाँ पद 78 है।
इसलिए a_n = 78
⇒ a + (n – 1)d = 78
⇒ 3 + (n – 1)(5) = 78
⇒ (n – 1)(5) = 75
⇒ n – 1 = 15
⇒ n = 16
अतः, A.P.: 3,8,13,18,… का 16वाँ पद 78 है।

निम्नलिखित समांतर श्रेढ़ी में कितने पद हैं? 7,13,19,…,205.

यहाँ, a = 7 तथा d = 13 – 7 = 6 है,
माना, समांतर श्रेढ़ी में n पद हैं।
इसलिए a_n = 205
⇒ a + (n – 1)d = 205
⇒ 7 + (n – 1)(6) = 205
⇒ (n – 1)(6) = 198
⇒ n – 1 = 33
⇒ n = 34
अतः, समांतर श्रेढ़ी में 34 पद हैं।

Write first four terms of the AP, when the first term a and the common difference d are given as follows: a = 10, d = 10.

a = 10, d = 10
First term a_1 = a = 10
Second term a_2 = a_1 + d = 10 + 10 = 20
Third term a_3 = a_2 + d = 20 + 10 = 30
Fourth term a_4 = a_3 + d = 30 + 10 = 40

For the following AP, write the first term and the common difference: 3,1,-1,-3,…

First term a = 3
Common difference d= a_2 – a_1 = 1 – 3 = – 2

उस A.P का 31वाँ पद ज्ञात कीजिए, जिसका 11वाँ पद 38 है और 16वाँ पद 73 है।

यहाँ, दिया है a_11 = 38 तथा a_16 = 73 है, ज्ञात करना है: a_31
दिया है: a_11= a + (11 – 1)d = 38
⇒ a + 10d = 38
⇒ a = 38 – 10d … (1)
तथा a_16 = 73
⇒ a + 15d = 73
समीकरण (1) से a का मान रखने पर
38 – 10d + 15d = 73
⇒ 5d = 35
⇒ d = 7
समीकरण (1) में d का मान रखने पर
a = 38 – 10(7) = -32
इसलिए, a_31 = a + 30d = -32 + 30(7) = 178
अतः, A.P का 31वाँ पद 178 है।

यदि किसी A.P. के तीसरे और नौवें पद क्रमशः 4 और -8 हैं, तो इसका कौन-सा पद शून्य होगा?

यहाँ, दिया है a_3 = 4 तथा a_9 = -8 है,
ज्ञात करना है: n, जहाँ a_n = 0.
दिया है: a_3 = a + (3 – 1)d = 4
⇒ a + 2d = 4
⇒ a = 4 – 2d … (1)
तथा a_9 = -8
⇒a + 8d = -8
समीकरण (1) से a का मान रखने पर
4 – 2d + 8d = -8
⇒ 6d = -12
⇒d = -2
समीकरण (1) में d का मान रखने पर
a = 4 – 2(-2) = 8
इसलिए, a_n = 0 में मान रखने पर
a_n = a + (n – 1) d = 0
⇒ 8 + (n – 1)(-2) = 0
⇒ n – 1 = 4
⇒ n = 5
अतः, इस A.P का 5वाँ पद शून्य होगा।

किसी A.P. का 17वाँ पद उसके 10वें पद से 7 अधिक है। इसका सार्व अंतर ज्ञात कीजिए।

माना, पहला पद = a तथा सार्व अंतर = d
प्रश्नानुसार, a_17 = a_10 + 7
⇒ a + 16d = a + 9d + 7
⇒ 7d = 7
⇒ d = 1
अतः, इस A.P का सार्व अंतर 1 है।

वह A.P. ज्ञात कीजिए जिसका तीसरा पद 16 है और 7वाँ पद 5वें पद से 12 अधिक है।

माना, समांतर श्रेणी का पहला पद = a तथा सार्व अंतर = d
तीसरा पद = 16
⇒ a_3 = 16
⇒ a + 2d = 16 … (1)
7वाँ पद 5वें पद से 12 अधिक है। इसलिए, a_7 = a_5 + 12
⇒ a + 6d = a + 4d + 12
⇒ 2d = 12
⇒ d = 6
समीकरण (1) में d का मान रखने पर, a + 2(6) = 16
⇒ a = 4
अतः, A.P. = a, a+d, a+2d,… = 4, 10, 16,…

सुब्बा राव ने 1995 में ₹5000 के मासिक वेतन पर कार्य आरंभ किया और प्रत्येक वर्ष ₹200 की वेतन वृद्धि प्राप्त की। किस वर्ष में उनका वेतन ₹7000 हो गया?

आरंभिक मासिक वेतन = a = ₹5000
प्रत्येक वर्ष वेतन वृद्धि (सार्व अंतर) = d = ₹200
माना, n वर्ष में उनका वेतन ₹7000 हो गया।
इसलिए, a_n = 7000
⇒ a + (n – 1)d = 7000
⇒ 5000 + (n – 1)(200) = 7000
⇒ (n – 1)(200) = 2000
⇒ n – 1 = 10
⇒ n = 11
अतः, 11वें वर्ष में उनका वेतन ₹7000 हो गया।