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

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

Toolbox:
• $n^{th}$ 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}$$(l+a) where l=last term and a= First term. Sum natural numbers lying between 100 and 1000 that are multiples of 5 is 105+110+115+..............995 Clearly this series is an A.P. with First\: term\:a=105,\:common\:difference\:d=5, and last\: term\:t_n=995 We know that t_n=a+(n-1)d \Rightarrow\:995=105+(n-1)5 \Rightarrow\:n=179 We know that S_n=\large\frac{n}{2}$$(l+a)$
$\therefore$ The sum of the required series $S_n=\large\frac{179}{2}$$(995+105)$
$=179\times 550=98450$