Question

If there are 2n + 1 terms in an arithmetic series, then prove that the ratio of the sum of odd terms to the sum of even terms is $n+1:n$.

Answer 1

Hint: We will break the AP into two APs, one having the odd terms and the other having even terms. Then we will find the first term and the common difference of the two series and then we will use the formula $\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ for the sum of series in AP and then we will find the ratio.

Complete step-by-step solution -
Let’s start solving this question.
Let the first term be ‘a’ which is the first odd term and let ‘d’ be the common difference.
$\begin{align}
  & {{T}_{1}}=a \\
 & {{T}_{2}}=a+d \\
 & {{T}_{3}}=a+2d \\
 & {{T}_{4}}=a+3d \\
\end{align}$
Similarly we can find the values of all the terms.
Now we can see that the odd terms are ${{T}_{1}},{{T}_{3}},{{T}_{5}}......$
Therefore, the first term in a series of odd terms is ‘a’, and the common difference is ${{T}_{3}}-{{T}_{1}}=a+2d-a=2d$, hence it is ‘2d’.
Now the even terms are ${{T}_{2}},{{T}_{4}},{{T}_{6}}.........$
Therefore, the first term in a series of even terms is ‘a+d’ and the common difference is ${{T}_{4}}-{{T}_{2}}=a+3d-a-d=2d$, hence it is ‘2d’.
Now the first even term will be ‘a + d’
It has been given that there are 2n + 1 terms in total.
Hence, there are n + 1 odd term and n even terms because the total number of terms is odd and the first term is also odd, so odd terms will be more.
Now for the series of odd terms,
The first term is ‘a’ and the common difference will be ‘2d’ because for difference ‘d’ we will get the even term after that.
Now the formula for sum of n terms of AP having first term ‘a’ and common difference ‘d’ is:
$\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right).........(1)$
Now using (1) for first term ‘a’, total terms n + 1, and the common difference ‘2d’ we get,
$\begin{align}
  & =\dfrac{n+1}{2}\left( 2a+\left( n+1-1 \right)2d \right) \\
 & =\dfrac{n+1}{2}\left( 2a+2nd \right)............(2) \\
\end{align}$
Now for the series of even terms,
We have first term as ‘a + d’, common difference as ‘2d’ and total terms n.
Hence, substituting these values is (1) we get,
 $\begin{align}
  & =\dfrac{n}{2}\left( 2(a+d)+\left( n-1 \right)2d \right) \\
 & =\dfrac{n}{2}\left( 2a+2d+2nd-2d \right) \\
 & =\dfrac{n}{2}\left( 2a+2nd \right)............(3) \\
\end{align}$
Now the ratio of sum of odd terms by even terms is:
$\begin{align}
  & =\dfrac{equation(2)}{equation(3)} \\
 & =\dfrac{\dfrac{n+1}{2}\left( 2a+2nd \right)}{\dfrac{n}{2}\left( 2a+2nd \right)} \\
 & =\dfrac{n+1}{n} \\
\end{align}$
Hence, proved.

Note: We have formed the two different AP to solve this question. If the total number of terms is odd then the number of the odd terms is one greater than the number of even term, as the first term is an odd term. One can check the relation by putting some value of n, a, and d to check whether it is true or not. The formula $\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ for the sum of AP of n terms must be kept in mind. The method that we used to divide one AP into sub-AP’s is a very important step.