Question

The ratio of the sums of first m and first n terms of an arithmetic series is ${{m}^{2}}:{{n}^{2}}$, prove that the ratio of ${{m}^{th}}$ and ${{n}^{th}}$ is $2m-1:2n-1$.

Answer 1

Hint: First we will look at the definition of arithmetic series and then we will write the formula for the sum of that series up to n terms and then another formula of it’s ${{n}^{th}}$ term. From the given ratio we will find the value of ‘d’ or common difference and then using that information we will find the ratio of two terms.

Complete step-by-step solution -
Arithmetic series: It is a series of numbers in which if we pick any two consecutive numbers then their difference will always be the same.
Now we will write the two formulas that we are going to use:
Sum up to ${{n}^{th}}$ term: $\dfrac{n\left( 2a+\left( n-1 \right)d \right)}{2}$
${{n}^{th}}$ term: $a+\left( n-1 \right)d$
In these two formulas ‘ a ‘ is the first term of the series and ‘ d ‘ is the common difference.
Now as per given in the question:
$\begin{align}
  & \dfrac{{{m}^{2}}}{{{n}^{2}}}=\dfrac{\dfrac{m\left( 2a+\left( m-1 \right)d \right)}{2}}{\dfrac{n\left( 2a+\left( n-1 \right)d \right)}{2}} \\
 & \dfrac{{{m}^{2}}}{{{n}^{2}}}=\dfrac{m\left( 2a+\left( m-1 \right)d \right)}{n\left( 2a+\left( n-1 \right)d \right)} \\
 & \\
\end{align}$
Now canceling m and n from both sides we get,
$\begin{align}
  & \dfrac{m}{n}=\dfrac{\left( 2a+\left( m-1 \right)d \right)}{\left( 2a+\left( n-1 \right)d \right)} \\
 & 2am+nmd-md=2an+nmd-nd \\
 & 2a\left( m-n \right)=\left( m-n \right)d \\
 & 2a=d \\
\end{align}$
Now that we have a relation between ‘ a ‘ and ‘ d ‘ so now we can find the ratio of ${{m}^{th}}$ and ${{n}^{th}}$ terms.
$\dfrac{{{m}^{th}}\text{ term}}{{{n}^{th}}\text{ term}}=\dfrac{a+\left( m-1 \right)d}{a+\left( n-1 \right)d}$
Now using the relation between ‘ a ‘ and ‘ d ‘ we get,
$\begin{align}
  & \dfrac{{{m}^{th}}\text{ term}}{{{n}^{th}}\text{ term}}=\dfrac{a+\left( m-1 \right)2a}{a+\left( n-1 \right)2a} \\
 & \dfrac{{{m}^{th}}\text{ term}}{{{n}^{th}}\text{ term}}=\dfrac{1+\left( m-1 \right)2}{1+\left( n-1 \right)2} \\
 & \dfrac{{{m}^{th}}\text{ term}}{{{n}^{th}}\text{ term}}=\dfrac{2m-1}{2n-1} \\
\end{align}$
Hence, we have found the ratio.
Hence proved.

Note: We can also solve this question by taking some values of ‘ a ‘ and ‘ d ‘ and from that we have a arithmetic series, now we just have use the formula for the sum and ${{n}^{th}}$ term and if we want we can also put any values of m and n and then solve this question.