Question

A geometric series consists of four terms and has a positive common ratio. The sum of the first two terms is $9$ and sum of the last two terms is $36$. Find the series

Answer 1

Let $a,ar,a{r}^2,a{r}^3$ be the first four terms of the given geometric series.

It is given that the sum of first two terms is $9$ that is:

$a+ar=9\Rightarrow a(1+r)=\mathrm{9.......}\left(1\right)$

It is given that the sum of last two terms is $36$ that is:

${ar}^2+{ar}^3=36\Rightarrow {ar}^2(1+r)=36\Rightarrow r^2\left(9\right)=36\left(using\right(1\left)\right)\Rightarrow 9r^2=36$

$\Rightarrow r^2=\frac{36}{9}\Rightarrow r^2=4\Rightarrow r=\pm \sqrt{4}\Rightarrow r=\pm 2$

The common ratio cannot be negative, thus substitute $r=2$ in equation 1 as follows:

$a(1+2)=9\Rightarrow 3a=9\Rightarrow a=\frac{9}{3}=3$

Now, with $a=3$ and $r=2$ then the first four terms of geometric series are:

$a=3$

$ar=3\times 2=6$

$a{r}^2=3\times 2^2=3\times 4=12$ and

$a{r}^3=3\times 2^3=3\times 8=24$

Hence, the series is $3+6+12+24+....$