Question

The $1$st term of an infinite G.P. exceeds the $2$nd term by $2$ and its sum is $50$. Find the series.

Answer 1

Let a be the first term & be r be the common ratio of the G.P

then the G.P will be

a,ar,$a{r}^2----$

It is given that a-ar=2

$a(1-r)=2\Rightarrow 1-r\frac{2}{a}--\left(1\right)$

And the sum is 50

$\Rightarrow \frac{a}{1-r}=50$

put $1-r=\frac{2}{a}form\left(1\right)$

$\frac{a}{\frac{2}{a}}=50$

$\frac{a^2}{2}=50$

$=a^2=100$

$\Rightarrow a=\pm 10$

with a =$\pm 10$

$\frac{10}{1-r}=50$

$\frac{1}{5}1-r$

$r=1-\frac{1}{5}=\frac{4}{5}$

$\Rightarrow$ series will be $10,8,\frac{32}{5}----$

with a = -10

$\frac{-10}{1-r}=50\Rightarrow \frac{-10}{50}=1-r=\frac{-1}{5}=1-r$ $r=1+\frac{1}{5}=\frac{6}{5}$

series will be$-10,-12,\frac{-72}{5}-----$