Question

The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is \[\dfrac{{27}}{{19}}\] . Then the common ratio of this series is:
$ {\text{A}}{\text{. }}\dfrac{{\text{4}}}{{\text{9}}} \\
  {\text{B}}{\text{. }}\dfrac{{\text{2}}}{{\text{9}}} \\
  {\text{C}}{\text{. }}\dfrac{{\text{2}}}{{\text{3}}} \\
  {\text{D}}{\text{. }}\dfrac{{\text{1}}}{{\text{3}}} $

Answer 1

Hint: Assume the two series and also assume their first term and the common ratio. Then use the formula for an infinite G.P. Form two equations and solve. We use $\dfrac{a}{{1 - r}}$ , where a is the first term and r is the common ratio.

Complete step-by-step answer:
Let the first term of the series be a and its common ratio is r.
Now, the series looks like $….. a, ar, ar^2, ar^3…..$
For this infinite G.P. the formula of sum becomes:
(1)……$\dfrac{a}{{1 - r}} = 3$ as stated in the question.
One more condition is given in the question which is the sum of the cubes of the terms of the first G.P.
The new G.P. becomes $…..a^3, a^3 r^3, a^3r^6, a^3r^9……..$
For this infinite series the sum of infinite series formula looks like..
(2)…..$\dfrac{{{a^3}}}{{1 - {r^3}}} = \dfrac{{27}}{{19}}$
Cubing the first equation (1)
$\dfrac{{{a^3}}}{{{{(1 - r)}^3}}} = 27$
Now dividing the two equations…(1)/(2)…we get
$\dfrac{{{{\left( {1 - r} \right)}^3}}}{{\left( {1 - {r^3}} \right)}} = \dfrac{1}{{19}}$
Now we can write
$\dfrac{{\left( {1 - r} \right){{\left( {1 - r} \right)}^2}}}{{\left( {1 - r} \right)\left( {1 + {r^2} + r} \right)}} = \dfrac{1}{{19}}$
$\left( {\because \left( {1 - {r^3}} \right) = \left( {1 - r} \right)\left( {1 + r + {r^2}} \right)} \right)$
$ {\text{19(1 - r}}{{\text{)}}^{\text{2}}}{\text{ = (1 + }}{{\text{r}}^{\text{2}}}{\text{ + r)}} \\
  \Rightarrow {\text{18}}{{\text{r}}^{\text{2}}}{\text{ - 39r + 18 = 0}} \\
 \Rightarrow {\text{6}}{{\text{r}}^{\text{2}}}{\text{ - 13r + 6 = 0}} \\
 \Rightarrow {\text{6}}{{\text{r}}^{\text{2}}}{\text{ - 9r - 4r + 6 = 0}} \\
\Rightarrow {\text{(2r - 3)(3r - 2) = 0}} \\
\Rightarrow {\text{r = }}\dfrac{{\text{2}}}{{\text{3}}}{\text{,}}\dfrac{{\text{3}}}{{\text{2}}} \\
\Rightarrow \because {\text{ |r| < 1, r = }}\dfrac{{\text{2}}}{{\text{3}}} $
The correct option is (C).

Note: To do this question, one must know the formula for an infinite G.P. and then it is just a simple calculative question. and also remember
$\left( {\because \left( {1 - {r^3}} \right) = \left( {1 - r} \right)\left( {1 + r + {r^2}} \right)} \right)$ these formula which helps in solving this type of questions easily.