Question

If the sum of $n$ terms of a G.P., with common ratio r, beginning with the $pth$ terms is $k$ times the sum of an equal number of terms of the same series beginning with the $qth$ terms, then the value of $k$ is _____.

• A. $\frac{rp}{q}$
• B. $\frac{rq}{p}$
• C. $r^{p-q}$
• D. $r^{p+q}$

Answer 1

Answer: (C)

$r^{p-q}$

Solution

option (C)$R^{p-q}$

Given,

No. of terms of G.P. is $n$

common ratio of G.P. is $r$

First term of G.P is starting from $P^{th}$ term

Let, first term of original G.P be $a$

So, The new G.P series starting with $p^{th}$ term is

$a{r}^{p-1},ar,......,a{r}^{n+p-1}$

sum of above G.P is

$S_{n_1}=\frac{a{r}^{p-1}(r^n-1)}{r-1}$ [$\because$ surn of G.P is $=\frac{a(r^n-1)}{r-1}]$

First term of G.P is starting from $q^{th}$ term

SDo, The new G.P series starting with $q^{th}$ term is $a{r}^{q-1},a{r}^q,a{r}^{q+1},......a{r}^{n+q-1}$

Sum of above G.P is

$S_{n_2}=\frac{a{r}^{q-1}(r^n-1)}{r-1}$

Given,

$S_{n_1}=K{S}_{n_2}$

$\frac{a{r}^{p-1}(r^n-1)}{r-1}=\frac{Ka{r}^{q-1}(r^n-1)}{r-1}$

$r^{p-1}=K{r}^{q-1}$

$\therefore K=\frac{r^{p-1}}{r^{q-1}}=r^{p-1-q+1}=r^{p-q}$

$K=r^{p-q}$