Question

The sum of an infinite G.P. is 2 and the sum of G.P. made from the cubes of this infinite G.P. is 24. Find the G.P.

Answer 1

Hint: In order to solve this problem you need to use the formula of the sum of infinite terms of the GP and then you have to make equations according to the question provided. Then you will get the required GP.

Complete step-by-step answer:
It is given that the sum of an infinite G.P. is 2.
Let the first term of the GP be a and the common ratio be r.
We know the sum of infinite GP is $\dfrac{{\text{a}}}{{{\text{1 - r}}}}$.
Then according to the above statement we can write the new equation as:
$\dfrac{{\text{a}}}{{{\text{1 - r}}}}$=2 …………..(1)
It is also given that the sum of G.P. made from the cubes of this infinite G.P. is 24.
So, its first term and common ratio will also be cubed.
Therefore the equation can be:
$\dfrac{{{{\text{a}}^3}}}{{{\text{1 - }}{{\text{r}}^3}}} = 24$ ……………………(2)
On cubing equation (1) and dividing it by equation (2) we get the new equation as:
$
  \dfrac{{{{\text{a}}^{\text{3}}}}}{{{{{\text{(1 - r)}}}^{\text{3}}}}}{\text{times }}\dfrac{{{\text{1 - }}{{\text{r}}^{\text{3}}}}}{{{{\text{a}}^{\text{3}}}}}{\text{ = }}\dfrac{{{{\text{2}}^{\text{3}}}}}{{{\text{24}}}}{\text{ = }}\dfrac{{\text{8}}}{{{\text{24}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{3}}} \\
  \dfrac{{{\text{1 - }}{{\text{r}}^{\text{3}}}}}{{{{{\text{(1 - r)}}}^{\text{3}}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{3}}} \\
$
And then further solving for common ratio we get,
$
  \dfrac{{{\text{1 + r + }}{{\text{r}}^{\text{2}}}}}{{{{{\text{(1 - r)}}}^{\text{2}}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{3}}}\,\,\,\,\,\,\,\,({\text{as (1 - r}}{{\text{)}}^3}{\text{ = (1 + r + }}{{\text{r}}^{\text{2}}}{\text{)(1 - r)}}\,\,{\text{and (1 - r}}{{\text{)}}^{\text{2}}}{\text{ = 1 + 2r + }}{{\text{r}}^{\text{2}}}{\text{)}} \\
  {\text{2}}{{\text{r}}^{\text{2}}}{\text{ + 5r + 2 = 0}} \\
  {\text{2}}{{\text{r}}^{\text{2}}}{\text{ + r + 4r + 2 = 0}} \\
  {\text{r(r + 2) + 2(2r + 1) = 0}} \\
  {\text{(r + 2)(2r + 1) = 0}} \\
  {\text{r = - }}\dfrac{{\text{1}}}{{\text{2}}}{\text{, - 2}} \\
$
So, on putting the value of common ratio r in equation (1) we get the equation as:
$
  \dfrac{{\text{a}}}{{{\text{1 - r}}}}{\text{ = 2}} \\
  \dfrac{{\text{a}}}{{{\text{1 - }}\left( {{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)}}{\text{ = 2}} \\
  {\text{a = 3}} \\
  {\text{and when we put r = - 2 we get, }} \\
  \dfrac{{\text{a}}}{{{\text{1 - ( - 2)}}}}{\text{ = 2}} \\
  {\text{a = (2)3}} \\
  {\text{a = 6}} \\
$
So, we get two different GPs with first term 6 and common ratio -2.
Second GP is with first term a = 3 and common ratio is $\dfrac{1}{2}$.
So the GPs are:
$ \Rightarrow $$3, - \dfrac{3}{2},\dfrac{3}{4}.............$
$ \Rightarrow $$6, - 12,24............$

Note: Whenever you face such types of problem you need to know that if you get the first term (a) and common ratio (r) then you can make the GP as ${\text{a,ar,a}}{{\text{r}}^{\text{2}}}{\text{,a}}{{\text{r}}^{\text{3}}}......$. Here we have made equations according to the question and solved the problem. Proceeding like this you will get the right solution.