Question

find the sum of integers between 100 and 200 which are divisible by 7

Answer 1

hope it helps you...

Answer 2

The sum of all the numbers between 100 and 200 which are divisible by 7 is 2107.

Given:

The numbers 100 and 200.

To Find:

The sum of integers between 100 and 200 which are divisible by 7.

Solution:

The numbers which are divisible by 7 between 100 and 200 are simply those numbers that occur after 7 places from the previous number.

Hence our sequence becomes 105, 112 ,119 ,126 ,..., 196.

Here,

The first term a = 105

The common difference d = 7

The general representation of the sequence is given by its term,

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

Therefore, $a_{n}$ = 105 + (n − 1)7 = 105 + 7n – 7 = 98 + 7n.

Now we need to find out the number of terms 'n' in the sequence.

The last term of the sequence $a_{n}$ = 196.

⇒ a + (n - 1)d = 196

⇒ 98 + 7n = 196

⇒ 7n = 98

⇒ n = 14

∴ There are 14 terms in the sequence.

Sum of n terms of an AP, $S_{n}$ = $\frac{n}{2}$ (2a + (n-1)d)

$S_{n}$ = $\frac{14}{2}$ (2x105 + (14 - 1)7 ) = 7(210 + 91) = 2107.

∴ The sum of all the numbers between 100 and 200 which are divisible by 7 is 2107.

#SPJ3