NCERT solutions for class 10 Maths chapter 5 exercise 5.4, 5.3, 5.2, 5.1 of AP – Arithmetic Progression 2018-2019 for UP Board (High Secondary) and CBSE Board in Hindi and English medium to download free in PDF form. * These Solutions* and

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## NCERT solutions for class 10 Maths chapter 5

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### Arithmetic Progression (AP) – NCERT solutions

##### English Medium

Hindi Medium solutions will be uploaded in May, 2018.

#### Previous Years Questions

##### Two marks questions

- Find how many integers between 200 and 500 are divisible by 8. [CBSE 2017]

##### Three marks questions

- Find the sum of n terms of the series (4 – 1/n) + (4 – 2/n) + (4 – 3/n) + …. [CBSE 2017]
- 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

- 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]
- 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.