Question

The sum of infinite GP is 57 and their cubes is 9747. Find the GP.

Answer 1

Hint: First apply the sum of geometric progression formula to the general sequence. Now find cubes of all terms. Try to observe the new sequence is a progression of type. After finding, apply the formula to this new sequence. Now you have 2 relations between 2 variables a, r. Solve these 2 equations by manipulating algebraically to find values of a, r. Thus getting the geometric progression, which is the required sequence.

Complete step-by-step solution -
Given condition in the question, is given by as follows:
The Sum of terms of Geometric progression is 57.
We know the sum of an infinite geometric progression of common ratio ‘r’ with the first term ‘a’ is given by $\dfrac{a}{{1 - r}}$.
Let us assume the first term of given progression is a, common ratio is r.
By substituting all values into the formula, we get it as:
$\dfrac{a}{{1 - r}}{\rm{ = 57 }}$ ……………….(1)
By applying cube on both sides of above equation, we get it as:
${\left( {\dfrac{a}{{1 - r}}} \right)^3}{\rm{ = 5}}{{\rm{7}}^3}$
By basic algebra knowledge, we can take the power n formula as:
${\left( {\dfrac{x}{y}} \right)^n}{\rm{ = }}\dfrac{{{x^n}}}{{{y^n}}}$
By substituting the above condition, we can get it in form of:
$\dfrac{{{a^3}}}{{{{\left( {1 - r} \right)}^3}}}{\rm{ = 5}}{{\rm{7}}^3}{\rm{ }}$ ……………….(2)
But our assumptions, we write the given progression as:
$a,{\rm{ ar, a}}{{\rm{r}}^2},{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{.}}$
By finding cube or each term, thus forming new sequence:
${\left( a \right)^3},{\rm{ }}{\left( {{\rm{ar}}} \right)^3}{\rm{, }}{\left( {{\rm{a}}{{\rm{r}}^2}} \right)^3},{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{.}}$
By basic properties of power, we have the formula as:
${\left( {xy} \right)^n}{\rm{ = }}{{\rm{x}}^{n{\rm{ }}}}{y^n};{\rm{ }}{\left( {{x^y}} \right)^n}{\rm{ = }}{{\rm{x}}^{yn}}$
By substituting these, we get the condition as follows:
${a^3},{\rm{ }}{{\rm{a}}^3}{r^3},{\rm{ }}{{\rm{a}}^3}{r^6},{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}$
First term $ = {\rm{ }}{{\rm{a}}^{\rm{3}}}$
Common ratio $ = \dfrac{{\text{second term}}}{{\text{first term}}} = {{\rm{r}}^{\rm{3}}}$
So, applying formula of sum to this new sequence, we get it as:
$\dfrac{{{a^3}}}{{1 - {r^3}}}{\rm{ = 9747 }}$ ……………….(3)
By dividing equation (2) with equation (3), we get it as:
$\dfrac{\dfrac{a^3}{(1-r)^3}}{\dfrac{a^3}{(1-r^3)}} = \dfrac{57^3}{9747} = \dfrac{57 \times 57 \times 57}{9747}$
We can write 9747 as 57 x 57 x 3, and cancelling common terms we get:
$\dfrac{{1 - {r^3}}}{{{{\left( {1 - r} \right)}^3}}}{\rm{ = }}\dfrac{{57 \times 57 \times 57}}{{57 \times 57 \times 3}}{\rm{ = }}\dfrac{{{\rm{57}}}}{3}{\rm{ = 19}}$
By writing \[1 - {r^3}{\rm{ = }}\left( {{\rm{1 - r}}} \right){\rm{ }}\left( {{\rm{1 + r + }}{{\rm{r}}^{\rm{2}}}} \right)\] we can cancel \[\left( {{\rm{1}} - {\rm{r}}} \right)\], we get:
\[\dfrac{{{\rm{1 + r + }}{{\rm{r}}^{\rm{2}}}}}{{{{\left( {1 - r} \right)}^2}}}{\rm{ = 19}}\]
By cross multiplying, we get the equation as:
\[{\rm{1 + r + }}{{\rm{r}}^{\rm{2}}}{\rm{ = 19}}{\left( {1 - r} \right)^2}\]
By expanding ${\left( {1 - r} \right)^2}$as\[{\rm{1 + }}{{\rm{r}}^{\rm{2}}}{\rm{ - 2r}}\], we get the equation as:
\[{\rm{1 + r + }}{{\rm{r}}^{\rm{2}}}{\rm{ = 19}}\left( {{r^2} - 2r + 1} \right)\]
By multiplying 19 inside the bracket on right hand side, we get:
\[{\rm{1 + r + }}{{\rm{r}}^{\rm{2}}}{\rm{ = 19}}{{\rm{r}}^{\rm{2}}}{\rm{ - 38r + 9}}\]
By subtracting $\left( {{r^2} + r{\rm{ + 1}}} \right)$ on both sides of equation, we get it as:
$19{r^2}{\rm{ + 19 - 38r - }}\left( {1{\rm{ + r + }}{{\rm{r}}^{\rm{2}}}} \right){\rm{ = 0}}$
By multiplying “-“inside the bracket, we get the equation as:
$19{r^2}{\rm{ + 19 - 38r - 1 - r - }}{{\rm{r}}^{\rm{2}}}{\rm{ = 0}}$
By combining the common terms, we get the equation as:
$19{r^2}{\rm{ - }}{{\rm{r}}^{\rm{2}}}{\rm{ - 38r - r + 19 - 1 = 0}}$
By simplifying the above equation, we get the equation as:
$18{r^2}{\rm{ - 39r + 18 = 0}}$
We can write 39r as $27r + 12r$, we get as
\[18{r^2}{\rm{ - 12r - 27r + 18 = 0}}\]
$ \Rightarrow \left( {3r - 2} \right)\left( {6r - 9} \right) = 0$
By above equation we get value of r to be as:
\[r{\rm{ = }}\dfrac{2}{3},{\rm{ }}\dfrac{9}{6}{\rm{ = }}\dfrac{2}{3},{\rm{ }}\dfrac{3}{2}\]
As we applied formula of \[{\rm{r}} < {\rm{1}}\], we take the value as \[{\rm{r}} = \dfrac{2}{3}\]
By substituting in equation (1), we get it as,
$\dfrac{a}{{1 - \dfrac{2}{3}}}{\rm{ = 57 }} \Rightarrow {\rm{ a = 57 }} \times {\rm{ }}\dfrac{1}{3}$
By simplification, we get the value of a as 19.
So, progression can be written as sequence below:
$19,{\rm{ 19 }} \times {\rm{ }}\dfrac{2}{3},{\rm{ 19 }} \times {\rm{ }}\dfrac{4}{9},{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{.}}$
By simplification, we get final progression as:
$19,{\rm{ }}\dfrac{{38}}{3},{\rm{ }}\dfrac{{76}}{9},{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}$

Note: Be careful while dividing equations, the idea of writing 9747 as \[{\rm{57 x 57 x 3}}\] is very important. While you get two roots of the equation remember to take only $\dfrac{2}{3}$ don’t confuse and take both you will get the wrong answer. To avoid confusion always verify the result you obtained.