NCERT Solutions for Class 10 Maths Chapter 5

NCERT solutions for class 10 Maths chapter 5 exercise 5.4, 5.3, 5.2, 5.1 of AP – Arithmetic Progression 2019-20 for UP Board (High Secondary) and CBSE Board in Hindi and English medium to download free in PDF form. Download Class 10 Apps based on updated NCERT Solutions for new academic session 2019-20.

 Class 10: Maths – गणित Chapter 5: Arithmetic Progression (AP) NCERT solutions for class 10 Maths chapter 5

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Arithmetic Progression (AP) – NCERT solutions

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Previous Years Questions

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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]

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.

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.

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.

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