Question

find the sum of all the numbers between 30 and 100 which are divisible by 11

Answer 1

hope this helps you ☺☺

Answer 2

Hey mate !!

Here's the answer !!

This question can be solved by using the concept of Arithmetic Progressions.

So the first number which is divisible by 11 is 33.

Hence a = 33 and d = 11.

We also know that $a_{n}$ = 99.

Hence we have to find " n " = ?

We know that,

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

So substituting the values we get,

99 = 33 + ( n - 1 ) 11

99 - 33 = 11 ( n - 1 )

66 = 11 ( n - 1 )

=> ( n - 1 ) = 66 / 11 = 6

=> n = 6 + 1 = 7

Hence the number of terms is 7.

So sum of all the seven terms can be arrived from the formula as follows:

$S_{n} = \frac{n}{2} [ { 2a + ( n - 1 ) d ]$

This implies that,

[tex]S_{7} = \frac{7}{2} [ 2 ( 33 ) + ( 7 - 1 ) 11 ] [/tex]

$S_{7} = \frac{7}{2} [ 66 + ( 6 ) 11 ]$

$S_{7} = \frac{7}{2} [ 66 + 66 ]$

$S_{7} = \frac{7}{2} * 132$

$S_{7} = 7 * 66$

$=\ \textgreater \ S_{7} = 462$

Hence the sum of all the numbers divisible by 11 from 30 to 100 is equal to 462.

Hope my answer helped !!

Cheers !!