Question

sum of 3 digit numbers divisible by 7

Answer 1

This is the answer for your question

Answer 2

Answer:

$Sum \:of \:3\:digit \: numbers\\divisible \:by \:7 = 70336$

Step-by-step explanation:

100,101,102,...,999 are 3 digit numbers.

105, 112, 119,....,994 are 3 digit numbers divisible by 7.

first term (a) = 105,

Last term (l) = 994

$common\: difference (d)\\=a_{2}-a_{1}\\=112-105\\=7$

$l = 994$

$\implies a+(n-1)d = 994$

$\implies 105+(n-1)7= 994$

/* Divide each term by 7, we get

$\implies 15+n-1 = 142$

$\implies14+n=142$

$\implies n = 142 - 14$

$\implies n = 128$

$Sum \:of \:n \:terms(S_{n})=\frac{n}{2}(a+l)$

$\implies S_{128}=\frac{128}{2}(105+994)$

$= 64\times 1099$

$=70336$

Therefore,

$Sum \:of \:3\:digit \: numbers\\divisible \:by \:7 = 70336$

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