Question

If the sum of positive terms of the series $10+9\frac{4}{7}+9\frac{1}{7}+.$ is $\frac{k}{7}$ then find value of $k$.

Answer 1

Answer: 852.00

$10+\frac{67}{7}+\frac{64}{7}+\dots \dots$
$=\frac{70}{7}+\frac{67}{7}+\frac{64}{7}\dots \dots +T_n$
let $T_n<0$
$\Rightarrow \frac{70}{7}+(n-1)\cdot \left(\frac{-3}{7}\right)<0$
$10-\frac{3}{7}n+\frac{3}{7}<0$
$\frac{3n}{7}>\frac{73}{7}$
$3n>73$
$n>24.3\dots \dots$
$\because n\in N$
$\Rightarrow n=25,26,\dots \dots$
$\Rightarrow T_{25}=$ first - ve term
$\therefore$ Sum of positive terms $=S_{24}$
$=\frac{24}{2}\left[2\times 10+(24-1)\left(\frac{-3}{7}\right)\right]$
$=12\times \left[20-\frac{69}{7}\right]$
$=\frac{12\times 71}{7}=\frac{852}{7}$