Question

Three positive numbers form an increasing GP. If the middle term in this GP is doubled then new numbers are in AP, find the common ratio of the GP. Again, if the middle term in the GP tripled, then check whether the common ratio of the GP will be same or not.

Or

Find the first term and common ratio of an infinite geometric series, if its sum is 4 and the second term is $\frac{3}{4}$.

Answer 1

Let a, ar and $a{r}^2$ be in GP.

On multiplying middle term by 2, i.e., a, 2ar and $a{r}^2$ are in AP.

$\Rightarrow 2\left(2ar\right)=a+a{r}^2$ $[\because a,bandcareinAP,so2b=1+c]$

$\Rightarrow 4r=1+r^2$

$\Rightarrow r^2-4r+1=0\Rightarrow r=\frac{4\pm \sqrt{16-4}}{2}$

$\Rightarrow r=2\pm \sqrt{3}\Rightarrow r=2+\sqrt{3}$ $[\because \text{GP is increasing, so we reject r=2-sqrt{3}}]$

Hence, common ratio of GP is $2+\sqrt{3}$.

If the middle term in the GP is tripled the a, 3ar and $a{r}^2$ are in AP. Then,

$2\left(3ar\right)=a+a{r}^2\Rightarrow r^2-6r+1=0$

$\therefore r=\frac{6\pm \sqrt{36-4}}{2}=\frac{6\pm \sqrt{32}}{2}=\frac{6\pm 4\sqrt{2}}{2}=3\pm 2\sqrt{3}$

Hence, common ratio will not be same.

Or

Let the first term be a and common ratio be r of an infinite geometric series.

We know that, $S_{\infty }=\frac{a}{1-r}\Rightarrow 4=\frac{a}{1-r}$

and $ar=\frac{3}{4}$

From eqs. (i) and (ii), we get

$\frac{a}{1-\frac{3}{4a}}=4\Rightarrow \frac{4a^2}{4a-3}=4$

$\Rightarrow 4a^2-16a+12=0\Rightarrow a^2-4a+3=0$

$\Rightarrow (a-1)(a-3)=0\Rightarrow a=1\text{or}a=3$

When a=1, $r=\frac{3}{4}$ and when $a=3,r=\frac{1}{4}$

$\therefore \text{First term be 1 or 3 and common ratio be}\frac{3}{4}\text{or}\frac{1}{4}\text{,~respectively}$.