Question

The first two terms of an infinite G.P are together equal to 5, and every term is 3 times the sum of all the terms that follow it. Find the series.

Answer 1

Hint: Write the sum of the first two terms that is $a+{{a}_{r}}=5$. Use ${{n}^{th}}$ term $=3\left[ {{\left( n+1 \right)}^{th}}\text{term}+{{\left( n+2 \right)}^{th}}\text{term}+{{\left( n+3 \right)}^{th}}\text{term}+.....\infty \right]$ i.e $3\left[ {{T}_{n+1}}+{{T}_{n+2}}+.....\infty \right]$ and put the terms by using the formula for the ${{n}^{th}}$ term of G.P that is $a{{r}^{n-1}}$.

Complete step-by-step answer:
Here, we are given that the sum of the first two terms of an infinite G.P is 5 and every term is 3 times the sum of all the terms that follow. We have to find this series.
We know that the geometric series is of the form
$a,ar,a{{r}^{2}},a{{r}^{3}}.....$
We are given that the sum of the first 2 terms of this G.P is 5.
Therefore, we get,
$a+ar=5....\left( i \right)$
Also, we are given that every term of the infinite G.P is 3 times the sum of all terms after it.
Therefore, we get
${{n}^{th}}\text{ term }=3\left[ {{\left( n+1 \right)}^{th}}\text{term}+{{\left( n+2 \right)}^{th}}\text{term}+{{\left( n+3 \right)}^{th}}\text{term}+.....\infty \right]$
We can write it as,
${{T}_{n}}=3\left[ {{T}_{n+1}}+{{T}_{n+2}}+{{T}_{n+3}}.....\infty \right]$
Since we know that ${{n}^{th}}$ term of G.P $=a{{r}^{n-1}}$, we get
$a{{r}^{n-1}}=3\left( a{{r}^{n+1-1}}+a{{r}^{n+2-1}}+a{{r}^{n+3-1}}....\infty \right)$
Or, $a{{r}^{n-1}}=3\left( a{{r}^{n}}+a{{r}^{n+1}}+a{{r}^{n+2}}+a{{r}^{n+3}}....\infty \right)$
By taking $a{{r}^{n}}$ common from both sides and canceling it, we get
$\Rightarrow {{r}^{-1}}=3\left( 1+{{r}^{1}}+{{r}^{2}}+{{r}^{3}}+{{r}^{4}}.....\infty \right)$
Since we know that the sum of the terms of infinite G.P with the first term as a and common ratio as $r =\dfrac{a}{1-r}$.
Therefore, we get
$\Rightarrow {{r}^{-1}}=3\left( \dfrac{1}{1-r} \right)$
As we know that $\left( {{p}^{-1}} \right)=\dfrac{1}{p}$, we get
$\Rightarrow \dfrac{1}{\left( r \right)}=\dfrac{3}{\left( 1-r \right)}$
By cross multiplying the above equation, we get
$\Rightarrow \left( 1-r \right)=3.r$
$\Rightarrow 3r+r=1$
$\Rightarrow 4r=1$
Therefore, we get $r=\dfrac{1}{4}$.
By putting the values of r in equation (i), we get
$\Rightarrow a+a\left( \dfrac{1}{4} \right)=5$
$\Rightarrow a+\dfrac{a}{4}=5$
$\Rightarrow \dfrac{5a}{4}=5$
Therefore, we get a = 4.
Therefore we get the first term as a = 4.
We get the second term as $ar=4.\dfrac{1}{4}=1$.
We get the third term as $a{{r}^{2}}=4.{{\left( \dfrac{1}{4} \right)}^{2}}=\dfrac{1}{4}$.
We get the fourth term as $a{{r}^{3}}=4.{{\left( \dfrac{1}{4} \right)}^{3}}=\dfrac{1}{16}$ and so on.
Therefore, we get the series as
$4,1,\dfrac{1}{4},\dfrac{1}{16},\dfrac{1}{64}.....\infty$

Note: Students should always check if the answer obtained is satisfying the conditions given in the question or not. Also, students must keep in mind that the sum of infinite G.P is $\dfrac{a}{1-r}$ while the sum of finite G.P is $\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}$ and read the questions properly to find which type of G.P is given in the question.