The sum of some terms of $G.P.$ is $315$ whose first term and the common ratio are $5$ and $2,$ respectively. Find the last term and the number of terms.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

Let the sum of n terms of the $G.P.$ be $315$

It is known that, $S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$

It is given that the first term $a$ is $5$ and common ratio $r$ is $2$

$\therefore 315=\frac{5\left(2^{n}-1\right)}{2-1}$

$\Rightarrow 2^{n}-1=63$

$\Rightarrow 2^{n}=64=(2)^{6}$

$\Rightarrow n=6$

$\therefore$ Last term of the $G.P.$ $=6^{\text {th }}$ term $=a r^{6-1}=(5)(2)^{5}=(5)(32)$

$=160$

Thus, the last term of the $G.P.$ is $160 .$

Similar Questions

If the ratio of the sum of first three terms and the sum of first six terms of a $G.P.$ be $125 : 152$, then the common ratio r is

If $1\, + \,\sin x\, + \,{\sin ^2}x\, + \,...\infty \, = \,4\, + \,2\sqrt 3 ,\,0\, < \,x\, < \,\pi $ then

Find the sum of the sequence $7,77,777,7777, \ldots$ to $n$ terms.

Let $S_1$ be the sum of areas of the squares whose sides are parallel to coordinate axes. Let $S_2$ be the sum of areas of the slanted squares as shown in the figure. Then, $\frac{S_1}{S_2}$ is equal to

  • [KVPY 2016]

If three successive terms of a$G.P.$ with common ratio $r(r>1)$ are the lengths of the sides of a triangle and $[\mathrm{r}]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to :

  • [JEE MAIN 2024]