Question

If the sum of an infinite GP is 3 and the sum of their squares is also 3. Then the first term and the common ratio of the series is ?
A) $1,\dfrac{1}{2}$
B) $\dfrac{3}{2},\dfrac{1}{2}$
C) $\dfrac{1}{{2,}}\dfrac{3}{2}$
D) None of these

Answer 1

Hint- Use the formula for the sum of the terms of an infinite GP on both the given GP and the GP of its squares. Square and divide both the equations using the given information to get the first term and common ratio.

Complete step-by-step answer:
Let the GP be $a,ar,a{r^2},a{r^3} \ldots \ldots \ldots \infty $, where a and r are the first term and common ratio respectively.
Now we know the sum of an infinite series
$ \Rightarrow $ $\dfrac{a}{{1 - r}} = 3$
On squaring,
$\dfrac{{{a^2}}}{{{{\left( {1 - r} \right)}^2}}} = {3^2}$
The square GP will be ${a^2},{a^2r^2},a^2{r^4} \ldots \ldots \ldots \infty $, where $ {a^2} $ and ${r^2}$ will be the first term and common ratio respectively.
We also know that the sum of the square G.P. is also 3.
$ \Rightarrow \dfrac{{{a^2}}}{{1 - {r^2}}} = 3$
On dividing both the equations
$\dfrac{{\dfrac{{{a^2}}}{{{{\left( {1 - r} \right)}^2}}}}}{{\dfrac{{{a^2}}}{{1 - {r^2}}}}} = \dfrac{9}{3} \Rightarrow \left( {\dfrac{{{1^2} - {r^2}}}{{{{\left( {1 - r} \right)}^2}}}} \right) = 3$
$ \Rightarrow \dfrac{{\left( {1 - r} \right)\left( {1 + r} \right)}}{{{{\left( {1 - r} \right)}^2}}} = 3$
$ \Rightarrow 1 + r = 3\left( {1 - r} \right) \Rightarrow 1 + r = 3 - 3r$
$ \Rightarrow $ $4r = 2 \Rightarrow r = \dfrac{1}{2}$
Since $\dfrac{a}{{1 - r}} = 3$ $ \Rightarrow a = 3\left( {1 - \dfrac{1}{2}} \right) \Rightarrow a = \dfrac{3}{2}$ ( putting value of r from above in the equation )
So option (B) is correct.
Note- In such types of questions, we should remember how to equate both the equations to find any one unknown variable and use it to find the other variable. Recall the formulas and concept of a GP to get to the required answer.