Question

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

Answer 1

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

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

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,

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

Answer 2

Answer:

hiii

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