Question

The sum of an infinitely decreasing GP is $4$ and the sum of cubes of its terms is equal to $\dfrac{64}{7}$. The ratio of its ${{5}^{th}}$ and ${{7}^{th}}$ term is…

Answer 1

Hint: In this question we have been given with the data that the sum of infinitely decreasing GP is $4$ and the sum of cubes of its terms is equal to $\dfrac{64}{7}$. And based on this data we have to find the ratio of the ${{5}^{th}}$ and ${{7}^{th}}$ term in the GP. GP stands for geometric progression and we will consider the infinite GP as $a+ar+a{{r}^{2}}+....\infty $, where $a$ is the first term, $r$ is the common multiple. We will make equations based on the given data and solve to get the required solution.

Complete step-by-step answer:
Let the geometric progression be as:
$\Rightarrow a+ar+a{{r}^{2}}+....\infty $
Now we know the sum of the GP is $4$ therefore, we can write:
$\Rightarrow a+ar+a{{r}^{2}}+....\infty =4\to \left( 1 \right)$
the sum of cubes of its terms is equal to $\dfrac{64}{7}$ therefore, on cubing the terms, we get:
$\Rightarrow {{a}^{3}}+{{a}^{3}}{{r}^{3}}+{{a}^{3}}{{r}^{6}}+....\infty =\dfrac{64}{7}\to \left( 2 \right)$
This makes the second GP have the first term as ${{a}^{3}}$ and the common multiple as ${{r}^{3}}$.
We know the formula that the sum of an infinite GP is $\dfrac{a}{1-r}$ given $0Therefore, we have the equations as:
$\Rightarrow \dfrac{a}{1-r}=4\to \left( 3 \right)$
 $\Rightarrow \dfrac{{{a}^{3}}}{1-{{r}^{3}}}=\dfrac{64}{7}\to \left( 4 \right)$
On cubing equation $\left( 3 \right)$, we get:
$\Rightarrow \dfrac{3{{a}^{3}}}{{{\left( 1-r \right)}^{3}}}=64\to \left( 5 \right)$
On dividing equation $\left( 4 \right)$ by $\left( 5 \right)$, we get:
$\Rightarrow \dfrac{\dfrac{{{a}^{3}}}{1-{{r}^{3}}}}{\dfrac{{{a}^{3}}}{{{\left( 1-r \right)}^{3}}}}=\dfrac{\dfrac{64}{7}}{64}$
Now on simplifying the right-hand side, we get:
$\Rightarrow \dfrac{\dfrac{{{a}^{3}}}{1-{{r}^{3}}}}{\dfrac{{{a}^{3}}}{{{\left( 1-r \right)}^{3}}}}=\dfrac{1}{7}$
On rearranging, we get:
$\Rightarrow \dfrac{{{\left( 1-r \right)}^{3}}}{1-{{r}^{3}}}=\dfrac{1}{7}$
Now on using the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+2ab+{{b}^{2}} \right)$ and expanding, we get:
$\Rightarrow \dfrac{{{\left( 1-r \right)}^{3}}}{\left( 1-r \right)\left( 1+2r+{{r}^{2}} \right)}=\dfrac{1}{7}$
On cancelling the terms, we get:
$\Rightarrow \dfrac{{{\left( 1-r \right)}^{2}}}{\left( 1+2r+{{r}^{2}} \right)}=\dfrac{1}{7}$
On using the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$, we get:
$\Rightarrow \dfrac{1-2r+{{r}^{2}}}{1+2r+{{r}^{2}}}=\dfrac{1}{7}$
On cross multiplying, we get:
$\Rightarrow 7+7{{r}^{2}}-14r=1+r+{{r}^{2}}$
On rearranging and simplifying, we get:
$\Rightarrow 6{{r}^{2}}-15r+6=0$
On splitting the middle term, we get:
$\Rightarrow 6{{r}^{2}}-12r-3r+6=0$
On taking the terms common, we get:
$\Rightarrow 6r(r-2)-3(r-2)=0$
On taking $\left( r-2 \right)$ common, we get:
$\Rightarrow \left( 6r-3 \right)\left( r-2 \right)$
We get $r=2$ or $r=\dfrac{1}{2}$. Since $r$ cannot be $2$since $0$\Rightarrow r=\dfrac{1}{2}$ as the value of $r$.
Now we have to find the ratio of$\dfrac{{{a}_{7}}}{{{a}_{5}}}$.
We can see from the sequence that the ${{7}^{th}}$ term is $a{{r}^{6}}$ and the ${{5}^{th}}$ term is $a{{r}^{4}}$ therefore, on substituting, we get:
$\Rightarrow \dfrac{{{a}_{7}}}{{{a}_{5}}}=\dfrac{a{{r}^{6}}}{a{{r}^{4}}}={{r}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}}=\dfrac{1}{4}$, which is the required solution.

Note: It is to be remembered that a geometric progression is said to be a decreasing geometric progression when the terms in the sequence are decreasing in value, this implies that the value of $r$ which is the common multiple is less than $1$. Other types of sequences should be remembered such as arithmetic progression.