# Arithmetic Progression

## NCERT Exercise 5.3

### Part 4

Question 4: How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

**Answer:** Here; a = 9, d = 8 and S_{n} = 636

We know;

`S=n/2[2a+(n-1)d]`

Or, `636=n/2[2xx9+(n-1)8]`

Or, `636=n(9+4n-4)`

Or, `n(4n+5)=636`

Or, `4n^2+5n=636`

Or, `4n^2+5n-636=0`

Or, `4n^2-48n+53n-636=0`

Or, `4n(n-12)+53(n-12)=0`

Or, `(4n+53)(n-12)=0`

Hence, `n=(53)/(4)` and `n=12`

Taking the integral value, we have n = 12

Question 5: 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.

**Answer:** We know;

`S=n/2[2a+(n-1)d]`

Or, `400=n/2[2xx5+(n-1)d]`

Or, `800=n(5+45)`

Or, `50n=800`

Or, `n=800÷50=16`

Now, common difference can be calculated as follows:

`a_n = a + (n – 1)d`

Or, `45 = 5 + 15d`

Or, `15d = 40`

Or, `d = (40)/(15) = 8/3`

Thus, n = 16 and `d = 8/3`

Question 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?

**Answer:** We have; a = 17, a_{n} = 350 and d = 9

We know;

`a_n = a + (n – 1)d`

Or, `350 = 17 + (n – 1)9`

Or, `(n – 1)9 = 350 – 17`

Or, `n – 1 = 333/9 = 37`

Or, `n = 38`

Now, sum can be calculated as follows:

`S=n/2[2a+(n-1)d]`

`=(38)/(2)(17+350)`

`=19xx367=6973`