Question

Find the sum of the integers between 100 and 200 that are divisible by 9.

Answer 1

108,117,126.... 198 and so on
So a=108
l=198
n= ?
nth term= a + (n-1)d
198= 108+(n-1)9
198= 108+9n-9
198-99= 9n
99/9= n
n= 11

Now, sum of n terms
Sum of first 11 terms= 11/2(108+198)
=11/2(306)
=11×153
=1683

Answer 2

Given:

"Integers" between "100" and "200"

To find:

The sum of the integers which is divisible by 9

Answer:

Given that

In between 100 to 200, the small number and that is divided by 9 is starting from 108

108, 117 … … … . 191, 198

Now,

$a = 108, d = 9\quad and \quad a_{n} = 198$

$a_{n} = a+ (n-1) \times d$

$198 = 108+ (n-1) 9$

$9 (n-1) = 198-108$

$9 (n-1) = 90$

$n-1 = \frac {90}{9}$

$n-1 = 10$

$n = 11$

$s = \frac {n \times (a_{1} + a_{n})} {2} ----- (1)$

Substitute n, a1 and an values in equation 1

$S = 11 \times \frac {108+198} {2}$

$= 11 times= \frac {306} {2}$

$= \frac {3366}{2}$

$S = 1683$