UPSEE Solved Model Paper 6

We can solve this problem using sum of n terms of an A.P.
We know that
The sum of n terms of an AP with first term 'a' and common difference 'd' is given by
Sn = n/2 (2a + (n-1)d)
By using this formula we can find the answer.
We have
The numbers which are divisible by 11 between 100 and 500 are 110, 121, ....., 495.
This is an A.P sequence.
First find number of terms using the formula
n =

last‌term‌first‌term

common‌difference

+1
Here,
last term = 495 , first term = 110
Common difference, d = 121 - 110 = 11
n =

495−110

11

+1
=

385

11

+1
= 35 + 1
= 36
S36 =

36

2

[2×110+(36−1)11]
[Since Sn =

n

2

(2a + (n - 1) d]
here, a = 110 , d = 11
= 18 [220 + 35 × 11]
= 18 [220 + 385]
= 18 × 605
= 10890
Therefore, the sum of all integers between 100 and 500 which are divisible by 11 is 108900