Find the sum to indicated number of terms in each of the geometric progressions in $\left.1,-a, a^{2},-a^{3}, \ldots n \text { terms (if } a \neq-1\right)$
Find the sum to indicated number of terms in each of the geometric progressions in $\left.1,-a, a^{2},-a^{3}, \ldots n \text { terms (if } a \neq-1\right)$
The given $G.P.$ is $1,-a, a^{2},-a^{3} \ldots \ldots$
Here, first term $=a_{1}=1$
Common ratio $=r=-a$
$S_{n}=\frac{a_{1}\left(1-r^{n}\right)}{1-r}$
$\therefore S_{n}=\frac{1\left[1-(-a)^{n}\right]}{1-(-a)}=\frac{\left[1-(-a)^{n}\right]}{1+a}$
Similar Questions
Find the sum of the products of the corresponding terms of the sequences $2,4,8,16,32$ and $128,32,8,2, \frac{1}{2}$
Find the sum of the products of the corresponding terms of the sequences $2,4,8,16,32$ and $128,32,8,2, \frac{1}{2}$
The sum of an infinite geometric series with positive terms is $3$ and the sum of the cubes of its terms is $\frac {27}{19}$. Then the common ratio of this series is
The sum of an infinite geometric series with positive terms is $3$ and the sum of the cubes of its terms is $\frac {27}{19}$. Then the common ratio of this series is
- [JEE MAIN 2019]
Let ${a_1},{a_2}...,{a_{10}}$ be a $G.P.$ If $\frac{{{a_3}}}{{{a_1}}} = 25,$ then $\frac {{{a_9}}}{{{a_{ 5}}}}$ equal
Let ${a_1},{a_2}...,{a_{10}}$ be a $G.P.$ If $\frac{{{a_3}}}{{{a_1}}} = 25,$ then $\frac {{{a_9}}}{{{a_{ 5}}}}$ equal
- [JEE MAIN 2019]
If every term of a $G.P.$ with positive terms is the sum of its two previous terms, then the common ratio of the series is
If every term of a $G.P.$ with positive terms is the sum of its two previous terms, then the common ratio of the series is
If $a, b, c$ and $d$ are in $G.P.$ show that:
$\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$
If $a, b, c$ and $d$ are in $G.P.$ show that:
$\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$