Question

If the sum of first $6$ terms is $9$ times to the sum of first $3$ terms of the same G.P., then common ratio of the series will be
$\left( 1 \right)$ $-2$
$\left( 2 \right)$ $2$
$\left( 3 \right)$ $1$
$\left( 4 \right)$ $\dfrac{1}{2}$

Answer 1

Hint: Since, for getting the sum of G.P, we need two things. One is the first term and second is the ratio. So, first of all we will assume first term and the ratio is $a$ and $r$ respectively. Then, we will find the sum of the first six terms and first three terms. Then, we will follow the given condition in the question and will simplify them to find the ratio.

Complete step-by-step solution:
Let us consider that the first term is $a$ and the ratio is $r$ . So, we need to find the value of $r$ .
Since, we have first term and ration, the series will be like as:
$\Rightarrow a,ar,a{{r}^{2}},a{{r}^{3}},...$
Here, we will suppose that $r$ is greater than $1$ . Then we will use the formula for getting the sum of G.P. up to $n$ terms as:
$\Rightarrow {{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$
Since, we need to find the sum of the first six terms. So, the sum of first six terms will be as:
$\Rightarrow {{S}_{6}}=\dfrac{a\left( {{r}^{6}}-1 \right)}{r-1}$
Now, we will find the sum of the first three terms of the same G.P. as:
$\Rightarrow {{S}_{3}}=\dfrac{a\left( {{r}^{3}}-1 \right)}{r-1}$
Since, in the question, given that the sum of first $6$ terms is $9$ times to the sum of first $3$ terms of the same G.P. So, according to the question:
$\Rightarrow {{S}_{6}}=9{{S}_{3}}$
Now, we will use the formula in the above step and will have the equation as:
$\Rightarrow \dfrac{a\left( {{r}^{6}}-1 \right)}{r-1}=9\dfrac{a\left( {{r}^{3}}-1 \right)}{r-1}$
Here, we can see that $a$ and $\left( r-1 \right)$ is a common factor in the above step. So, we can cancel out these two and will have the above equation as:
$\Rightarrow \left( {{r}^{6}}-1 \right)=9\left( {{r}^{3}}-1 \right)$
We can write $\left( {{r}^{6}}-1 \right)$ as $\left( {{\left( {{r}^{3}} \right)}^{2}}-{{1}^{2}} \right)$ in the above step below as:
$\Rightarrow \left( {{\left( {{r}^{3}} \right)}^{2}}-{{1}^{2}} \right)=9\left( {{r}^{3}}-1 \right)$
Now, we will use the formula of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the above step and the equation can be as:
$\Rightarrow \left( {{r}^{3}}+1 \right)\left( {{r}^{3}}-1 \right)=9\left( {{r}^{3}}-1 \right)$
Here, we will divide by $\left( {{r}^{3}}-1 \right)$ both sides in the above step and cancel out the term $\left( {{r}^{3}}-1 \right)$ as:
\[\Rightarrow \dfrac{\left( {{r}^{3}}+1 \right)\left( {{r}^{3}}-1 \right)}{\left( {{r}^{3}}-1 \right)}=\dfrac{9\left( {{r}^{3}}-1 \right)}{\left( {{r}^{3}}-1 \right)}\]
\[\Rightarrow \left( {{r}^{3}}+1 \right)=9\]
Now, we can write the number one side of the equal sign as:
\[\Rightarrow {{r}^{3}}=9-1\]
After solving above term, we will have:
\[\Rightarrow {{r}^{3}}=8\]
Since, $8$ is a cube of $2$ . So, we can write it as:
\[\Rightarrow {{r}^{3}}={{2}^{3}}\]
After comparing both, we will have the value as:
\[\Rightarrow r=2\]
Hence, the ratio of the series is $2$ .

Note: Here, we can check it by placing the value of $r$ in the given condition in the question as:
Since, the given condition is:
$\Rightarrow \dfrac{a\left( {{r}^{6}}-1 \right)}{r-1}=9\dfrac{a\left( {{r}^{3}}-1 \right)}{r-1}$
Now, we will put the value of $r$ as:
$\Rightarrow \dfrac{a\left( {{2}^{6}}-1 \right)}{2-1}=9\dfrac{a\left( {{2}^{3}}-1 \right)}{r-1}$
Now, we will solve denominator firs as:
$\Rightarrow \dfrac{a\left( {{2}^{6}}-1 \right)}{1}=9\dfrac{a\left( {{2}^{3}}-1 \right)}{1}$
So, we can write the above equation as:
$\Rightarrow a\left( {{2}^{6}}-1 \right)=9a\left( {{2}^{3}}-1 \right)$
Now, we will calculate the value of $2$ after simplifying power as:
$\Rightarrow a\left( 64-1 \right)=9a\left( 8-1 \right)$
Here, we will do subtraction as:
$\Rightarrow a\times 63=9a\times 7$
After multiplication, we will get:
$\Rightarrow 63a=63a$
Since, L.H.S = R.H.S.
Hence, the solution is correct.