# Arithmetic Progression

## NCERT Exercise 5.2

### Part 5

Question: 12 – Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

**Solution:** Since, both the APs have same common difference, thus in this case the difference between each corresponding terms will be 100.

Question: 13 – How many three digit numbers are divisible by 7?

**Solution:** Since, 100 is the smallest three digit number and it gives a reminder of 2 when divided by 7, therefore, 105 is the smallest three digit number which is divisible by 7.

Since, 999 is greatest three digit number, and it gives a reminder of 5, thus 999 – 5 = 994 will be the greatest three digit number which is divisible by 7.

Therefore, here we have,

First term (a) = 105,

The last term (a_{n}) = 994

The common difference = 7

We know that, `a_n = a + (n – 1)d`

Or, `994 = 105 + (n – 1)7`

Or, `(n – 1)7 = 994 – 105 = 889`

Or, `n – 1 = 127`

Or, `n = 128`

Thus, there are 128 three digit numbers which are divisible by 7.

Question: 14 – How many multiples of 4 lie between 10 and 250?

**Solution:** 12 is the first number after 10 which is divisible by 4.

Since, 250 gives a remainder of 2 when divided by 4, thus 250 – 2 = 248 is the greatest number less than 250 which is divisible by 4.

Here, we have first term (a) = 12, last term (n) = 248 and common difference (d) = 4

Thus, number of terms (n) =?

We know that, `a_n = a + (n -1)d`

Or, `248 = 12 + (n – 1)4`

Or, `(n – 1)4 = 248 – 12 = 236`

Or, `n – 1 = 59`

Or, `n = 60`

Thus, there are 60 numbers between 10 and 250 that are divisible by 4.

Question: 15 – For what value of n, are the nth terms of two APs; 63, 65, 67, ………. and 3, 10, 17, ……. equal.

**Solution:** In first AP: a = 63, d = 2

In second AP: a = 3, d = 7

As per question:

`63 + (n – 1) 2 = 3 + (n – 1) 7`

`⇒ 63 – 3 + (n – 1) 2 = (n – 1) 7`

`⇒ 60 + 2n – 2 = 7n – 7`

`⇒ 2n + 58 = 7n – 7`

`⇒ 2n + 58 + 7 = 7n`

`⇒ 2n + 65 = 7n`

`⇒ 7n – 2n = 65`

`⇒ 5n = 65`

`⇒ n = (65)/(5) = 13`

Thus, for the 13 value of n, nth term of given two APs will be equal