Question

Find the sum of the infinite A.G.P: 3, 4, 4, $\dfrac{32}{9}$, …
(a) 16
(b) 16
(c) 24
(d) 27

Answer 1

Hint: A general infinite A.G.P. can be written as the following: $a+\left( a+d \right)r+\left( a+2d \right){{r}^{2}}+...+\infty $. Compare this to the given series to find a = 3. Use the fact that the second and the third term are the same. So, $\left( 3+d \right)r=\left( 3+2d \right){{r}^{2}}=4$. Using this, find the value of r and d. Next, to find the summation of the A.G.P., multiply both sides by $\dfrac{2}{3}$ and subtract from the original equation. This will result in a G.P. Find the sum of this G.P. using \[S=\dfrac{a}{1-r}\]. Put this into the previous equation to get the final answer.

Complete step-by-step answer:
In this question, we are given the infinite A.G.P: 3, 4, 4, $\dfrac{32}{9}$, …
We need to find the sum of this infinite A.G.P.
First let us define what an A.G.P. actually is.
An arithmetic-geometric progression (A.G.P.) is a progression in which each term can be represented as the product of the terms of an arithmetic progressions (A.P.) and a geometric progressions (G.P.). 
A general infinite A.G.P. can be written as the following:
$a+\left( a+d \right)r+\left( a+2d \right){{r}^{2}}+...+\infty $
In this question, we have the first term, a = 3
Also, we see that the second and the third term are the same. So, using this information, we will get the following:
$\left( 3+d \right)r=\left( 3+2d \right){{r}^{2}}=4$ ….(1)
On solving this, we get the following:
$\dfrac{4}{r}-3=\left( \dfrac{4}{{{r}^{2}}}-3 \right)\cdot \dfrac{1}{2}$
$\left( 4-3r \right)\cdot 2r=4-3{{r}^{2}}$
$3{{r}^{2}}-8r+4=0$
Factorising this by splitting the middle term, we will get the following:
$\left( r-2 \right)\left( 3r-2 \right)=0$
$r=2,\dfrac{2}{3}$
But we need to find a finite sum till infinite terms. So, r cannot be greater than 1. Hence, r = 2 is rejected.
Hence, $r=\dfrac{2}{3}$
Substituting this in equation (1), we will get the following:
$\left( 3+d \right)r=4$
$\left( 3+d \right)\cdot \dfrac{2}{3}=4$
$3+d=6$
$d=3$
Now, we will find the sum of the infinite series.
$S=3+6\cdot \dfrac{2}{3}+9{{\left( \dfrac{2}{3} \right)}^{2}}+...\infty $
Multiply both sides by $\dfrac{2}{3}$, we will get the following:
$\dfrac{2}{3}S=3\cdot \dfrac{2}{3}+6{{\left( \dfrac{2}{3} \right)}^{2}}+9{{\left( \dfrac{2}{3} \right)}^{3}}+...\infty $
Subtracting these two equations, we will get the following:
\[S\left( 1-\dfrac{2}{3} \right)=3+3\cdot \dfrac{2}{3}+3{{\left( \dfrac{2}{3} \right)}^{2}}+...\infty \]
Now, this is in the form of a G.P. The sum of an infinite G.P. with r < 1 is given by:
\[S=\dfrac{a}{1-r}\]
Using this formula, we will get the following:
\[\dfrac{S}{3}=\dfrac{3}{1-\tfrac{2}{3}}=9\]
\[S=27\]
So, the sum of the infinite A.G.P: 3, 4, 4, $\dfrac{32}{9}$, … is 27.
Hence, option (d) is correct.

Note: In this question, it is very important to what an A.G.P is. Also, it is important to know that a general infinite A.G.P. can be written as the following: $a+\left( a+d \right)r+\left( a+2d \right){{r}^{2}}+...+\infty $. It is very important to eliminate r = 2 here because if r = 2, then the sum of the series would extend to infinity.