Find the sum of all integers between 101 and 500, which are divisible by 9.

Question:

Find the sum of all integers between 101 and 500, which are divisible by 9.

 

Solution:

To Find: Sum of all integers between 101 and 500 divisible by 9

The integers between 101 and 500 divisible by 9 are $108,117,126, \ldots, 495$ (Add 9 to 108 to get 117,9 to 117 to get 126 and so on).

Let a be the first term and d be the common difference and n be the number of terms of the AP

Here a = 108, d = 9, l = 495

$\Rightarrow a+(n-1) d=495$

$\Rightarrow 108+9(n-1)=495$

$\Rightarrow 12+(n-1)=55$

$\Rightarrow n=55-11=44$

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

$\Rightarrow S=\frac{44}{2}[2 \times 108+(44-1) 9]$

⇒ S = 22[216 + 387] = 22[603] = 13266

Sum of all integers divisible by 9 between 100 and 500 is 13266.

 

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