Question

Find the sum of the integers between 100 and 500 that are divisible by 9.​

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

The first integer between 100 and 500 divisible by 9 is :- 108.

The last integer between 100 and 500 divisible by 9 is :- 495.

The AP will be :- 108,117......495.

The formula for the sum of all the integers is :-

$\huge\boxed{\tt{\dfrac {n}{2}(2a+(n-1)d) }}$

So,

• a = 108
• d = 117 - 108 = 9
• n = ?

To find the sum, firstly we to find n.

To find n, we will apply the formula :- $\tt {a_n=a+(n-1)d}$.

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

$\tt\implies 495 = 108 + 9n - 9$

$\tt\implies 9n - 9 = 495 - 108$

$\tt\implies 9n - 9 = 387$

$\tt\implies 9n = 396$

$\tt\implies n = 44$

Using this in the formula of n,

$\tt\leadsto S_n=\dfrac {44}{2}(2(108)+(44-1)9)$

$\tt\leadsto S_n=\dfrac {44}{2}(2(108)+(44-1)9)$

$\tt\leadsto S_n=\dfrac {44}{2}(2(108)+(43)9)$

$\tt\leadsto S_n=22(216+387)$

$\tt\leadsto S_n=22(603)$

$\tt\leadsto S_n = 13266$

Therefore, the Sum of the integers between 100 and 500 that are divisible by 9 is :-

$\huge\boxed {\tt {S_n = 13266}}$

Answer 2

$\bf{\Huge{\boxed{\tt{\purple{ANSWER\::}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

The integers between 100 & 500 that are divisible by 9.

$\bf{\Large{\underline{\bf{To\:Find\::}}}}$

The sum.

$\bf{\Large{\underline{\tt{\pink{Explanation\::}}}}}$

The integers between 100 & 500 divisible by 9  are 108, 117, 126.........,495.

$\bf{Let\:the\:Arithmetic\:progression}\begin{cases}\sf{First\:term\:be\:=\:(a)}\\ \sf{Common\:difference\:=\:(d)}\\ \sf{Number\:of\:terms\:=\:(n)}\\ \sf{Last\:term\:=\:(l)}\end{cases}}$

We have,

• a = 108
• d = 9
• l = 495

According to the question :

$\implies\sf{an\:=\:a+(n-1)d}$

$\implies\sf{495\:=\:108+(n-1)9}$

$\implies\sf{495\:=\:108+9n-9}$

$\implies\sf{495\:=\:9n+99}$

$\implies\sf{9n\:=\:495-99}$

$\implies\sf{9n\:=\:396}$

$\implies\sf{n\:=\:\cancel{\frac{396}{9} }}$

$\implies\sf{\red{n\:=\:44}}$

__________________________________

We know that formula of the sum of last term, we get;

$\bf{\Large{\boxed{\sf{Sn\:=\:\frac{n}{2} (a+l)}}}}}$

So,

$\longmapsto\sf{Sn\:=\:\cancel{\frac{44}{2}} (108+495)}$

$\longmapsto\sf{Sn\:=\:22(108+495)}$

$\longmapsto\sf{Sn\:=\:22(603)}$

$\longmapsto\sf{Sn\:=\:(22*603)}$

$\longmapsto\sf{\red{Sn\:=\:13266}}$

Thus,

$\bf{\Large{\boxed{\bigstar{\rm{The\:sum\:of\:an\:A.P.\:is\:13266}}}}}$