Question

If the sum of an infinite G.P be $ 9 $ and the sum of first two terms be $ 5 $ , then the common ratio is
A. $ \dfrac{1}{3} $
B. $ \dfrac{3}{2} $
C. $ \dfrac{3}{4} $
D. $ \dfrac{2}{3} $

Answer 1

Hint: First, we shall analyze the given information so that we are able to solve the problem. Here, we are given the sum of an infinite G.P and the sum of the first two terms. We call a sequence a geometric progression when each succeeding term is obtained by multiplying each preceding with the constant value. That constant value is called the common ratio of G.P, and $ a $ is the first term of G.P and $ r $ is the common ratio of G.P. We are asked to calculate the common ratio.
Formula to be used:
The formula to obtain the sum of an infinite G.P is as follows.
The sum of an infinite G.P, $ S = \dfrac{a}{{1 - r}} $ where $ a $ is the first term of G.P and $ r $ is the common ratio of G.P.

Complete step by step answer:
It is given that the sum of an infinite G.P is $ 9 $ .
We shall apply the formula $ S = \dfrac{a}{{1 - r}} $ here.
Then, we get $ 9 = \dfrac{a}{{1 - r}} $
 $ \Rightarrow 9\left( {1 - r} \right) = a $
 $ \Rightarrow 9 - 9r = a $
 $ \Rightarrow a = 9 - 9r $ ……….. $ \left( 1 \right) $
Also, we are given that the sum of the first two terms is $ 5 $
The first two terms of G.P will be $ a,ar $
Hence, we get $ a + ar = 5 $
\[ \Rightarrow ar = 5 - a\]
\[ \Rightarrow r = \dfrac{{5 - a}}{a}\] ……… $ \left( 2 \right) $
Now, we shall substitute $ \left( 2 \right) $ in $ \left( 1 \right) $ .
 $ \Rightarrow a = 9 - 9\left( {\dfrac{{5 - a}}{a}} \right) $
 $ \Rightarrow a = \dfrac{{9a - 45 + 9a}}{a} $
 $ \Rightarrow {a^2} = 9a - 45 + 9a $
 $ \Rightarrow {a^2} - 18a + 45 = 0 $
We shall split the middle term.
 $ \Rightarrow {a^2} - 15a - 3a + 45 = 0 $
 $ \Rightarrow a\left( {a - 15} \right) - 3\left( {a - 15} \right) = 0 $
 $ \Rightarrow \left( {a - 3} \right)\left( {a - 15} \right) = 0 $
Hence we get $ a = 3 $ or $ a = 15 $ .
Now, from $ \left( 2 \right) $ we get
\[ \Rightarrow r = \dfrac{{5 - 3}}{3}orr = \dfrac{{5 - 15}}{{15}}\]
\[ \Rightarrow r = \dfrac{2}{3}\] or\[r = \dfrac{{ - 12}}{{15}}\]
Here \[r = \dfrac{2}{3}\] is possible.

So, the correct answer is “Option D”.

Note: The common ratio will be negative only if the given geometric progression contains alternate positive and negative terms. Since there are not any details about the alternate terms, we choose\[r = \dfrac{2}{3}\].
    Also, if we are given the sum of $ {n^{th}} $ terms of G.P, then we need to apply the formula $ {S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}},r > 1 $ and $ {S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}},r \ne 1 $ where $ a $ is the first term of G.P, $ r $ is the common ratio of G.P and is $ n $ the number of terms.