Question

Sum of infinite terms $A,AR,A{R}^2,A{R}^3,\dots$ is $15$. If the sum of the squares of these terms is $150$, then find the sum of $A{R}^2,A{R}^4,A{R}^6,\dots$.

Answer 1

Step-1 : Applying the formula of the sum of an infinite geometric series

Formula to be used : We know that the sum of an infinite geometric series $a,ar,a{r}^2,a{r}^3,\cdots$ with common ratio $r$ is $\frac{a}{1-r}$.

Here, the given series is $A,AR,A{R}^2,A{R}^3,\dots$which is an infinite geometric series with common ratio $R$.. Then its sum will be $\frac{A}{1-R}$.

Now, by the question, $\frac{A}{1-R}=15$. $\left(1\right)$

Step-2 : Finding the sum of the squares of the terms of the series $A,AR,A{R}^2,A{R}^3,\dots$

After squaring each term, we get the series as : $A^2,A^2{R}^2,A^2{R}^4,A^2{R}^6,\dots$ which is also an infinite geometric series with common ratio $R^2$. So, its sum will be $\frac{A^2}{1-R^2}$. Now, by the question, $\frac{A^2}{1-R^2}=150$. $\left(2\right)$

Step-3 : Finding the value of $A$

From Step-1 and Step-2, we get :

$$\begin{gathered} \frac{A^2}{1-R^2}=150 \\ \Rightarrow \frac{A.A}{\left(1-R\right)\left(1+R\right)}=150\left[\because \left(1-R^2\right)=\left(1+R\right)\left(1-R\right)\right] \\ \Rightarrow \frac{A}{1-R}.\frac{A}{1+R}=150 \\ \Rightarrow \frac{A}{1+R}=\frac{150}{15}\left[\because \frac{A}{1-R}=15\right] \\ \Rightarrow A=10\left(1+R\right) \end{gathered}$$

Now, Substituting $A=10\left(1+R\right)$ in $\left(1\right)$, we obtain :

$$\begin{gathered} \frac{A}{1-R}=15 \\ \Rightarrow \frac{10\left(1+R\right)}{1-R}=15 \\ \Rightarrow \frac{1+R}{1-R}=\frac{3}{2} \\ \Rightarrow 3-3R=2+2R \\ \Rightarrow 5R=1 \\ \therefore R=\frac{1}{5} \end{gathered}$$

Hence,

$$\begin{gathered} A=10\left(1+R\right) \\ =10\left(1+\frac{1}{5}\right) \\ =10\times \frac{6}{5} \\ =12 \end{gathered}$$

Step-4 : Finding the sum of $A{R}^2,A{R}^4,A{R}^6,\dots$

The series $A{R}^2,A{R}^4,A{R}^6,\dots$ is an infinite geometric series with common ratio $R^2$. So, its sum will be

$$\begin{gathered} \frac{A{R}^2}{1-R^2}=\frac{12.{\left({\displaystyle \frac{1}{5}}\right)}^2}{1-{\left({\displaystyle \frac{1}{5}}\right)}^2} \\ =\frac{12}{24} \\ =\frac{1}{2} \end{gathered}$$

Hence, the sum of $A{R}^2,A{R}^4,A{R}^6,\dots =\frac{1}{2}$.