Question

In an infinite G.P., the sum of first three terms is 70. If the extreme terms are multiplied by 4 and the middle term is multiplied by 5, the resulting terms form an A.P. then the sum to infinite terms of G.P.
A. 120
B. – 40
C. 160
D. 80

Answer 1

Hint: First of all, consider the first three terms of the G.P. as variables then use the condition of the terms to be in A.P. after multiplying the extreme and middle terms with the given constants. So, use this concept to reach the solution of the given problem.

Complete step by step solution:
Let the first three terms of the G.P. be \[a,ar,a{r^2}\].
Given that their sum is equal to 70 i.e.,
\[
   \Rightarrow a + ar + a{r^2} = 70 \\
  \therefore a\left( {1 + r + {r^2}} \right) = 70...............................\left( 1 \right) \\
\]
Also, given that the extreme terms are multiplied by 4 and the middle term is multiplied by 5. Then the terms are given by \[4a,5ar,4a{r^2}\]. And these terms are in A.P.
We know that the condition to be in A.P. for the three terms \[x,y,z\] is given by \[2y = x + z\].
Since, \[4a,5ar,4a{r^2}\] are in A.P. and by the condition to be in A.P. we have
\[
   \Rightarrow 2\left( {5ar} \right) = 4a + 4a{r^2} \\
   \Rightarrow 10ar = 4a + 4a{r^2} \\
   \Rightarrow 2a\left( {5r} \right) = 2a\left( {2 + 2{r^2}} \right) \\
\]
Cancelling the common terms, we have
\[
   \Rightarrow 5r = 2 + 2{r^2} \\
   \Rightarrow 2{r^2} - 5r + 2 = 0 \\
   \Rightarrow 2{r^2} - 4r - r + 2 = 0 \\
   \Rightarrow 2r\left( {r - 2} \right) - 1\left( {r - 2} \right) = 0 \\
   \Rightarrow \left( {2r - 1} \right)\left( {r - 2} \right) = 0 \\
  \therefore r = \dfrac{1}{2},2 \\
\]
We know that for an infinite term of G.P. the common ratio must be less than one i.e., \[r < 1\].
So, \[r = \dfrac{1}{2}\].
Substituting \[r = \dfrac{1}{2}\] in equation \[\left( 1 \right)\], we have
\[
   \Rightarrow a\left( {1 + \dfrac{1}{2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) = 70 \\
   \Rightarrow a\left( {1 + \dfrac{1}{2} + \dfrac{1}{4}} \right) = 70 \\
   \Rightarrow a\left( {\dfrac{{4 + 2 + 1}}{4}} \right) = 70 \\
   \Rightarrow a\left( {\dfrac{7}{4}} \right) = 70 \\
   \Rightarrow a = 70 \times \dfrac{4}{7} \\
  \therefore a = 40 \\
\]
We know that the sum of infinite terms in G.P. with the first term \[a\] and common ratio \[r\] is given by \[{S_\infty } = \dfrac{a}{{1 - r}}\].
Hence, the sum of the infinite terms of G.P. is \[{S_\infty } = \dfrac{{40}}{{1 - \dfrac{1}{2}}} = \dfrac{{40}}{{\dfrac{1}{2}}} = 40 \times 2 = 80\]
Thus, the correct option is D. 80
Note: For an infinite term of G.P. the common ratio must be less than one i.e., \[r < 1\] and their sum is given by \[{S_\infty } = \dfrac{a}{{1 - r}}\]. The extreme terms in a sequence are the first and last terms of that sequence.