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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- $n^{th}$ 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)$

$=179\times 550=98450$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...