Question

If the sum of an infinite GP is 20 and sum of their square is 100 then ratio will be
\[\begin{align}
  & A.\dfrac{1}{2} \\
 & B.\dfrac{1}{4} \\
 & C.\dfrac{3}{5} \\
 & D.1 \\
\end{align}\]

Answer 1

Hint: In this question, we are given the sum of infinite GP and the sum of their square. We need to find the common ratio. For this, we will first suppose as the first term and r as the common ratio and then form a geometric progression using them. Then we will use the formula of a sum of infinite terms in a GP given by $ S=\dfrac{a}{1-r} $ to form an equation. Then we will square those terms and again find the sum and form an equation. Solving both equations will give us the value of a and r. 'r' will be the required common ratio.

Complete step by step answer:
Here let us suppose that the first term of an geometric progression as a and the common ratio as r. Our geometric progression will then look like this, $ a,ar,a{{r}^{2}},a{{r}^{3}},\ldots \ldots \infty $ .
( $ a,ar,a{{r}^{2}},a{{r}^{3}},\ldots \ldots $ are in GP with common ratio r).
We are given the sum of infinite terms in the geometric progression as 20.
Since sum of any infinite term in a geometric progression with A as first term and R as common ratio is given by $ S=\dfrac{A}{1-R} $ . So for formed geometric progression $ S=\dfrac{a}{1-r} $ we can say that,
 $ \dfrac{a}{1-r}=20\cdots \cdots \cdots \left( 1 \right) $ .
Now each term of the GP is squared, so new GP becomes equal to $ a,ar,a{{r}^{2}},a{{r}^{3}},\ldots \ldots \infty $ .
For this geometric progression, first term becomes $ {{a}^{2}} $ and common ratio becomes $ {{r}^{2}} $ (Because $ \dfrac{{{a}^{2}}{{r}^{2}}}{{{a}^{2}}}={{r}^{2}}=\dfrac{{{a}^{2}}{{r}^{4}}}{{{a}^{2}}{{r}^{2}}} $ )
Sum of this new GP is given as 100, so we can say $ \dfrac{{{a}^{2}}}{1-{{r}^{2}}}=100\cdots \cdots \cdots \left( 2 \right) $ .
To simplify our calculations let us square equation (1), squaring both sides in equation (1) we get,
 $ \begin{align}
  & \dfrac{{{a}^{2}}}{{{\left( 1-r \right)}^{2}}}={{\left( 20 \right)}^{2}} \\
 & \Rightarrow \dfrac{{{a}^{2}}}{{{\left( 1-r \right)}^{2}}}=400\cdots \cdots \cdots \left( 3 \right) \\
\end{align} $ .
Now let us solve equations (2) and (3) to get the value of r. Dividing equation (2) by equation (3) we get:
 $ \dfrac{\dfrac{{{a}^{2}}}{1-{{r}^{2}}}}{\dfrac{{{a}^{2}}}{{{\left( 1-r \right)}^{2}}}}=\dfrac{400}{100} $ .
Rearranging the left side and solving the right side we get,
 $ \dfrac{{{a}^{2}}}{1-{{r}^{2}}}\times \dfrac{{{\left( 1-r \right)}^{2}}}{{{a}^{2}}}=4 $
Cancelling $ {{a}^{2}} $ from the numerator and the denominator of the left side of the equation, we get,
 $ \dfrac{{{\left( 1-r \right)}^{2}}}{1-{{r}^{2}}}=4 $ .
We know that $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ so writing $ 1-{{r}^{2}} $ as $ {{1}^{2}}-{{r}^{2}} $ and then using the property we get,
 $ \dfrac{{{\left( 1-r \right)}^{2}}}{\left( 1-r \right)\left( 1+r \right)}=4 $ .
Canceling one of the (1-r) from the numerator by (1-r) from the denominator we get,
 $ \dfrac{1-r}{1+r}=4 $ .
Cross multiplying we get,
\[1-r=4\left( 1+r \right)\Rightarrow 1-r=4+4r\].
Rearranging the terms we get,
 $ 4r+r=4-1\Rightarrow 5r=3 $ .
Dividing both sides by 5 we get,
 $ r=\dfrac{3}{5} $ .
Hence the required common ratio is $ \dfrac{3}{5} $ .
Hence option C is the correct answer.

Note:
Students should take care of the signs while solving this sum. Make sure to change values of common ratio also while forming the new geometric progression. Students can also find the value of a by putting the value of r in equation (1).