Question

Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7

Answer 1

I hope it's help you...

Answer 2

Answer:

sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7 = 19668

Explanation:

First integer after 100 which on division with 16 leaves 7 as

remainder is 103

103, 119,135, ......,791 are in

A.P

first term (a) = 103

common difference (d) = a2-a1

= 119 - 103

d = 16

i) nth term = an = 791

We know that,

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

=> 103+(n-1)×16 = 791

=> (n-1)×16 = 791 - 103

=> (n-1)×16 = 688

=> n-1 = 688/16

=> n-1 = 43

=> n = 43+1

=> n = 44

ii ) we know that,

Sum of n terms in A.P = Sn

$\boxed { S_{n}= \frac{n}{2}\left(a+a_{n}\right)}$

$S_{44} = \frac{44}{2}\left(103+791\rifht)$

= $ 22\times 894$

= $ 19668$

Therefore,

sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7 is 19668

••••