Question

The sum of infinite G.P. is 23 and the sum of the squares of the series is 69. The first term is
A.$\dfrac{{69}}{{13}}$
B.$\dfrac{{10}}{3}$
C.$\dfrac{9}{{10}}$
D.$\dfrac{{13}}{{69}}$

Answer 1

Here we need to find the first term of the series. We will first assume the first term to be any variable and the common ratio to be any variable. Then we will use the formula of sum of the series of the GP. Then we will find the sum of squares for each term of the series using the same formula. After solving the equations, we will get the required answer.

Complete step-by-step answer:
Let the first term of the series to be $a$ and the common ratio be $r$.
According to question;
$a + ar + a{r^2} + a{r^3} + ...... = 23$ …………. $\left( 1 \right)$
We know the formula of sum GP is given by Sum $= \dfrac{a}{{1 - r}}$.
Using this formula in equation $\left( 1 \right)$, we get
$\Rightarrow \dfrac{a}{{1 - r}} = 23$ ……….. $\left( 2 \right)$
Now, we will square each term of the GP and equating it to the sum 69, we get
${a^2} + {a^2}{r^2} + {a^2}{r^4} + {a^2}{r^6} + ...... = 69$ ……….. $\left( 3 \right)$
Here we can see that the first term of this series is ${a^2}$ and the common ratio is ${r^2}$.
Now we will apply the formula of the sum of the GP.
$\Rightarrow \dfrac{{{a^2}}}{{1 - {r^2}}} = 69$
Using the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ in the denominator, we get
Using this identity in the denominator, we get
$\Rightarrow \dfrac{{{a^2}}}{{\left( {1 - r} \right)\left( {1 + r} \right)}} = 69$ …………… $\left( 4 \right)$
Now, we will square equation $\left( 2 \right)$ and then we will divide equation $\left( 4 \right)$ by equation $\left( 2 \right)$.
$\Rightarrow \dfrac{{\dfrac{{{a^2}}}{{\left( {1 - r} \right)\left( {1 + r} \right)}}}}{{\dfrac{{{a^2}}}{{{{\left( {1 - r} \right)}^2}}}}} = \dfrac{{69}}{{{{23}^2}}}$
On further simplifying the terms, we get
$\Rightarrow \dfrac{{1 - r}}{{1 + r}} = \dfrac{3}{{23}}$
On cross multiplying the terms, we get
$\Rightarrow 23 - 23r = 3 + 3r$
Now, we will add or subtract the like terms, we get
$\Rightarrow 20 = 26r$
Now, we will divide both sides by 26.
$\begin{array}{l} \Rightarrow \dfrac{{20}}{{26}} = \dfrac{{26r}}{{26}}\\\ \Rightarrow \dfrac{{10}}{{13}} = r\\\ \Rightarrow r = \dfrac{{10}}{{13}}\end{array}$
Now, we will substitute the value of common ratio in equation 2.
$\Rightarrow \dfrac{a}{{1 - \dfrac{{10}}{{13}}}} = 23$
On further simplifying the terms, we get
$\Rightarrow \dfrac{a}{{\dfrac{3}{{13}}}} = 23$
On cross multiplying the terms, we get
$\Rightarrow a = \dfrac{3}{{13}} \times 23$
On multiplying the numbers, we get
$\Rightarrow a = \dfrac{{69}}{{13}}$
Therefore, the first term of the given geometric series is equal to $\dfrac{{69}}{{13}}$.
Hence, the correct option is option A.

Note: Here we have obtained the first term of the GP. Here, GP stands for the Geometric Progression and it is defined as the series whose ratio between any consecutive terms is always a constant. We should not get confused between the geometric progression and arithmetic progression. An arithmetic progression is a sequence or series where there is a common difference between consecutive terms.