Question

find the sumnof 3 digit numbers which are multiples of 8

Answer 1

$\huge\purple{\underline{\underline{\pink{Ans}\red{wer:-}}}}$

$\sf{The \ sum \ of \ three \ digit \ numbers \ which}$

$\sf{are \ multiple \ of \ 8 \ is \ 61376.}$

$\sf\orange{To \ find:}$

$\sf{The \ sum \ of \ three \ digit \ numbers \ which}$

$\sf{are \ multiple \ of \ 8.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Three \ digit \ numbers \ which \ are \ multiple}$

$\sf{of \ 8 \ are:-}$

$\sf{104,112,120,...,992}$

$\sf{Here, \ t1=104, \ t2=112, \ t3=120}$

$\sf{t2-t1=112-104=8}$

$\sf{t3-t2=120-112=8}$

$\sf{\therefore{The \ sequence \ form \ is \ in \ A.P.}}$

$\sf{with \ common \ difference (d) \ as \ 8.}$

$\sf{In \ the \ A.P.}$

$\sf{104,112,120,…,992}$

$\sf{Here, \ t1=a=104, \ d=8 \ and \ tn=992}$

$\sf{tn=a+(n-1)d... formula}$

$\sf{\therefore{992=104+(n-1)8}}$

$\sf{8(n-1)=992-104}$

$\sf{8(n-1)=888}$

$\sf{n-1=\frac{888}{8}}$

$\sf{n-1=111}$

$\sf{n=111+1}$

$\sf{n=112}$

__________________________________

$\sf{Sn=\frac{n}{2}[t1+tn]…formula}$

$\sf{\therefore{S112=\frac{112}{2}[104+992]}}$

$\sf{S112=56\times1096}$

$\sf{S112=61376}$

$\sf\purple{\tt{\therefore{The \ sum \ of \ three \ digit \ numbers \ which}}}$

$\sf\purple{\tt{are \ multiple \ of \ 8 \ is \ 61376.}}$