Question

The sum of an infinite number of terms in G.P. is 23 and the sum of their squares is 69. Find the series.

Answer 1

Hint: In the question involving the concept of the geometric series, it is important to note that the sum of the terms in an infinite G.P. is given by, $\dfrac{a}{{1 - r}}$ , where $a$ is the first term and $r$ is the common ratio. Also, the sum of the terms of any infinite G.P. is given by,
$a + ar + a{r^2} + a{r^3} + .....$ . Thus, using the given conditions and applying the formula of sum of infinite G.P., you will get the required answer.


Complete step-by-step solution
Given that the sum of the infinite term of terms in G.P. is 23. Let us translate this into a mathematical equation:
$a + ar + a{r^2} + a{r^3} + ........ = 23{\text{ ......(1)}}$
Also, it is given to you that the sum of their squares is 69. Again, we translated this into a mathematical equation.
\[{a^2} + {\left( {ar} \right)^2} + {\left( {a{r^2}} \right)^2} + {\left( {a{r^3}} \right)^2} + ........ = 69\]
Let us simplify the obtained expression to get a clear picture of this condition.
\[{a^2} + {a^2}{r^2} + {a^2}{r^4} + {a^2}{r^6} + ........ = 69{\text{ ......(2)}}\]
The first term in the above series (on the left side) is ${a^2}$ and the common ratio is ${r^2}$ . Also, we know that the sum of the terms of the infinite series is given by, $\dfrac{a}{{1 - r}}$ , where $a$ is the first term and $r$ is the common ratio.
Use this formula to find the sum of the series(on the left) in (1) and (2) and simplify them.
The sum of the series(on the left) in (1) is given by;
 $
  \dfrac{a}{{1 - r}} = 23 \\
   \Rightarrow a = 23\left( {1 - r} \right){\text{ ......(3)}} \\
$
Next, apply this sum formula to simplify equation (2).
$
  \dfrac{{{a^2}}}{{1 - {r^2}}} = 69 \\
   \Rightarrow {a^2} = 69\left( {1 - {r^2}} \right){\text{ ......(4)}} \\
$

Now, let us solve equation (3) and (4) to get the unknown values.
\[
  {\left( {23\left( {1 - r} \right)} \right)^2} = 69\left( {1 - {r^2}} \right) \\
   \Rightarrow {23^2}{\left( {1 - r} \right)^2} = 69\left( {1 - r} \right)\left( {1 + r} \right) \\
   \Rightarrow 23\left( {1 - r} \right) = 3\left( {1 + r} \right) \\
   \Rightarrow 23 - 23r = 3 + 3r \\
   \Rightarrow 26r = 20 \\
   \Rightarrow r = \frac{{10}}{{13}} \\
\]
Now you have got the value of the common ratio r, all you need to do now is to substitute this obtained value of r in (3) to find the value of a.
\[
  a = 23\left( {1 - r} \right) \\
   \Rightarrow a = 23\left( {1 - \dfrac{{10}}{{13}}} \right) \\
   \Rightarrow a = 23\left( {\dfrac{3}{{13}}} \right) \\
   \Rightarrow a = \dfrac{{69}}{{13}} \\
\]

Now we know that the G.P. is given by; $a,ar,a{r^2},....$ . Thus, we substitute the value of ‘a’ and ‘r’ in this series to find the required G.P.
$\dfrac{{69}}{{13}},\left( {\dfrac{{69}}{{13}}} \right)\left( {\dfrac{{10}}{{13}}} \right),\left( {\dfrac{{69}}{{13}}} \right){\left( {\dfrac{{10}}{{13}}} \right)^2},.....$

Hence, we have obtained our required series.

Note: In order to get command over these types of questions, you need to have a grip over the formulas used in the concept of G.P. Also, you need to be familiar with how to calculate the common ratio. Knowing the formula and applying the given conditions to form the equations will lead you to your final answer. Also, please don’t confuse yourself between the common ratios of AP and GP, as one thing being wrong will make your entire solution wrong. Most importantly, take care of the calculation mistakes.