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Find the sum of all the natural numbers lying between $100\:\: and\:\: 1000$ which are multiples of $5$

$\begin{array}{1 1}98450 \\ 97900 \\ 99550 \\ 99450 \end{array} $

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  • $n^{th}$ term of an A.P., $t_n=a+(n-1)d$ where $a=$First term and $d=$common difference
  • Sum of $n$ terms of an A.P., $S_n=\large\frac{n}{2}$$(l+a)$ where $l$=last term and $a=$ First term.
Sum natural numbers lying between $100$ and $1000 that are multiples of 5 is
Clearly this series is an $A.P.$ with
$First\: term\:a=105,\:common\:difference\:d=5,$ and $last\: term\:t_n=995$
We know that $t_n=a+(n-1)d$
We know that $S_n=\large\frac{n}{2}$$(l+a)$
$\therefore$ The sum of the required series $S_n=\large\frac{179}{2}$$(995+105)$
$=179\times 550=98450$
answered Feb 19, 2014 by rvidyagovindarajan_1

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