Question

If the sum of the first 2n, 3n, and 5n terms of an A.P. be ${{S}_{2}},{{S}_{3}},{{S}_{5}}$ respectively. Find the value of ${{S}_{3}}-{{S}_{2}}$ in terms of ${{S}_{5}}$.

Answer 1

Hint: We first try to find the general equations and series for an A.P. We assume the first term and the common difference. We find the equation of general term and sum of first n terms of the series. Then we put the values of 2n, 3n, and 5n in place of the equations and find the relation between ${{S}_{2}},{{S}_{3}},{{S}_{5}}$.

Complete step-by-step solution
The sum of the first 2n, 3n, and 5n terms of an A.P. be ${{S}_{2}},{{S}_{3}},{{S}_{5}}$.
We express the A.P. in its general form.
We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
Let’s assume the first term is a. So, ${{t}_{1}}=a$. The common difference is d.
We express general term ${{t}_{n}}$ as ${{t}_{n}}={{t}_{1}}+\left( n-1 \right)d=a+\left( n-1 \right)d$.
The sum of n terms will be ${{S}_{n}}$ as ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.
To place the conditional value of 2n, 3n and 5n we have to place the values in ${{S}_{n}}$.
We replace values of 2n, 3n, and 5n in place of n.
${{S}_{2}}=\dfrac{2n}{2}\left[ 2a+\left( 2n-1 \right)d \right]=n\left[ 2a+\left( 2n-1 \right)d \right]$
${{S}_{3}}=\dfrac{3n}{2}\left[ 2a+\left( 3n-1 \right)d \right]$
${{S}_{5}}=\dfrac{5n}{2}\left[ 2a+\left( 5n-1 \right)d \right]$
Now we need to find the value of ${{S}_{3}}-{{S}_{2}}$.
So, ${{S}_{3}}-{{S}_{2}}=\dfrac{3n}{2}\left[ 2a+\left( 3n-1 \right)d \right]-n\left[ 2a+\left( 2n-1 \right)d \right]$
We solve the equation to get
$\begin{align}
  & {{S}_{3}}-{{S}_{2}}=\dfrac{3n}{2}\left[ 2a+\left( 3n-1 \right)d \right]-n\left[ 2a+\left( 2n-1 \right)d \right] \\
 & =2a\left( \dfrac{3n}{2}-n \right)+dn\left( \dfrac{9n}{2}-2n \right)-d\left( \dfrac{3n}{2}-n \right) \\
 & =an+\dfrac{5d{{n}^{2}}}{2}-\dfrac{dn}{2} \\
\end{align}$
Now we need to express ${{S}_{3}}-{{S}_{2}}$ in terms of ${{S}_{5}}$.
We have ${{S}_{5}}=\dfrac{5n}{2}\left[ 2a+\left( 5n-1 \right)d \right]=5\left( an+\dfrac{5d{{n}^{2}}}{2}-\dfrac{dn}{2} \right)$
We replace the value to get
$\begin{align}
  & {{S}_{5}}=5\left( an+\dfrac{5d{{n}^{2}}}{2}-\dfrac{dn}{2} \right)=5\left( {{S}_{3}}-{{S}_{2}} \right) \\
 & \Rightarrow {{S}_{3}}-{{S}_{2}}=\dfrac{{{S}_{5}}}{5} \\
\end{align}$
So, the value of ${{S}_{3}}-{{S}_{2}}$ in terms of ${{S}_{5}}$ is ${{S}_{3}}-{{S}_{2}}=\dfrac{{{S}_{5}}}{5}$.

Note: We need to remember that the general equation was also made of n. when we are replacing the value with 2n, 3n, and 5n we can’t confuse the n of the equations. We are just placing a value and if that becomes too confusing then we can start with r instead of n.