Question

Find the sum of an infinite decreasing geometric progression if the sum of the first four terms is equal to 15 and the sum of their squares is equal 85.

Answer 1

Let first term be a and common ratio$ r.$
$\therefore a\left(1+r+r^2+r^3\right)=15$....(1)
$\text { and } a^2\left(1+r^2+r^4+t^6\right)=85$....(2)
Now, squaring the first equation and dividing the result by the second equation, we get
$\frac{\left(1+r+r^2+r^3\right)^2}{\left(1+r^2+r^4+t^6\right)}=\frac{45}{17} \Rightarrow \frac{\left(\frac{r^4-1}{r-1}\right)^2}{\left(\frac{r^8-1}{r^2-1}\right)}=\frac{\left(r^4-1\right)^2}{(r-1)^2} \cdot \frac{\left(r^2-1\right)}{\left(r^8-1\right)}=\frac{\left(r^4-1\right)}{(r-1)} \cdot \frac{(r+1)}{\left(r^4+1\right)} $
$=\frac{\left(r^3+r^2+r+1\right)(r+1)}{\left(r^4+1\right)}=\frac{r^4+2 r^3+2 r^2+2 r+1}{r^4+1}=\frac{45}{17} \Rightarrow 14 r^4-17 r^3-17 r^2-17 r+14=0$
As $r=0$ is not a root of above equation, so we get (deleting by $r ^2$ )
$14\left( r ^2+\frac{1}{ r ^2}\right)-17\left( r +\frac{1}{ r }\right)-17=0$

$\therefore r+\frac{1}{r}=\frac{5}{2} \Rightarrow r=2, \frac{1}{2}$
Note that $r+\frac{1}{r}=\frac{-9}{7}$ (Rejected)
So, for $r =\frac{1}{2}$, from equation (1), we get $a =8$
$\therefore $ Terms of infinite decreasing geometric progression are $8,4,2,1, \frac{1}{2}$.........
Hence, $S _{\infty}=\frac{8}{1-\frac{1}{2}}=\frac{8}{\frac{1}{2}}=16$