Question

find the sum of all multiples of 3 natural number between 1 and hundred​

Answer 1

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Answer 2

Answer:

$\boxed{=1683}$

Step-by-step explanation:

$\text{The multiples of 3 are 3,6,9,12,..........99} \ \ \ [upto \ 100] \\\\ The \ series \ are \ in \ AP. \\\\ first \ term,a=3 \\ common \ difference,d=6-3=9-6=3 \\\\n^{th} \ term,a_n=99\\\\ => a+(n-1)d=99 \\ 3+(n-1)(3)=99\\ (n-1)(3)=99-3\\(n-1)(3)=96\\n-1=96/3\\n-1=32\\n=32+1\\n=33\\\\ total \ no.of \ terms,n=33\\\\ =>Sum \ of \ n \ terms \ is \ given \ by,\\ S_n=\frac{n}{2} [2a+(n-1)d]\\\\ S_{33}=\frac{33}{2}[2(3)+(33-1)(3)]\\\\S_{33}=\frac{33}{2}[6+96]\\\\ S_{33}=\frac{33}{2}[102]\\\\ S_{33}=33\times51$

$S_{33}=1683$

the sum of multiples of 3 between 1 and 100 is 1683