Question

If $S_1,S_2,S_3,,,,,S_p$ are the sums of infinite geometric series, whose first digits are $1,2,3,..,p$ respectively and whose common ratios are $\frac{1}{2},\frac{1}{3},\frac{1}{4},......\frac{1}{p+1}$ respectively then $s_1+s_2+s_3+......+s_p$ is

• A. $\frac{p}{2}(p+4)$
• B. $\frac{p}{2}(p+3)$
• C. $\frac{p}{2}(p+1)$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{p}{2}(p+3)$

$\begin{array}{c}{S}_1,S_2.S_3......S_{p}areinite\\ G.P\\ firstterm=1,2,3,\mathrm{4.....}p\\ commonratio=\frac{1}{2},\frac{1}{3},\frac{1}{4}....\frac{1}{p}\end{array}$

sum of infinite GP $\frac{a}{L\cdot r}$

$\begin{array}{c}{S}_1=\frac{1}{1-\frac{1}{2}}=2\\ S_2=\frac{2}{1-\frac{1}{3}}=3\\ S_3=\frac{3}{1-\frac{1}{4}}=4\\ S_p=\frac{p}{1-\frac{1}{p+1}}=p+1\\ hencesum=S_1+S_2+S_3+......+S_p\\ =2+3+4+.....+\left(p+1\right)\\ S_n=\frac{p}{2}\left[2a+\left(p-1\right)1\right]\\ =\frac{p}{2}\left[4+p-1\right]\\ \frac{p\left(p+3\right)}{2}\end{array}$