Question

The nth term of a series is given to be $\frac{3+n}{4},$ find the sum of 105 term of this series.

Answer 1

nth term of $+ve$ series $=\frac{3+n}{4}$

So, the series becomes,

$1,\frac{5}{4},\frac{6}{4},\frac{7}{4}$.......

$\therefore a=1,d=\frac{5}{4}-1=\frac{1}{4}=\frac{6}{4}-\frac{5}{4}$

$\therefore S_{105}=\frac{105}{2}[2\times 1+(105-1\left)\frac{1}{4}\right]$

$=\frac{105}{2}[2+104\times \frac{1}{4}]$

$=\frac{105}{2}\left[\frac{8+104}{4}\right]$

$=\frac{105}{2}\times \frac{112}{4}$

$=105\times 14$

$=1470$

$\therefore S_{105=1470}$