Question

If the sum of first m terms of an arithmetic progression is n and the sum of first n terms of that arithmetic progression is m then show that the sum of first m + n terms of that arithmetic progression is – (m + n).

Answer 1

Hint: This question requires the formula of first n terms of an arithmetic progression when its first term and the common difference is given. Let the first term of an arithmetic progression be a and the common difference be d. Then the arithmetic progression is
a, a + d, a + 2d, a + 3d, . . . , a + (n-1)d

The sum of the first n term is given by $S=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$ .


Complete step-by-step answer:

We are given that the sum of first m terms of an arithmetic progression is n. So it means that we have to use the formula of sum of arithmetic progression which is given below.
If a is the first term of an arithmetic progression and d is the common difference then the sum of first n term is given by
$S=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$
So applying this formula we get that
Let the first term of our given AP be a and its common difference be d. Then
$n=\dfrac{m}{2}\left[ 2a+(m-1)d \right]\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(i)}$
It is also given that the sum of first n terms is m. So
$m=\dfrac{n}{2}\left[ 2a+(n-1)d \right]\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(ii)}$
Subtracting the previous two equations we get,
$\begin{align}
  & n-m=ma-na+\dfrac{m(m-1)}{2}d-\dfrac{n(n-1)}{2}d \\
 & n-m=\left( m-n \right)a+\dfrac{{{m}^{2}}-m-{{n}^{2}}+n}{2}d \\
 & n-m=\left( m-n \right)a+\dfrac{\left( m-n \right)\left( m+n-1 \right)}{2}d \\
\end{align}$
Cancelling (m-n) from both side,
$\begin{align}
  & -1=a+\dfrac{m+n-1}{2}d \\
 & \Rightarrow (m+n-1)d=-2a-2\,\,\,\,\,\,\,\,\,\,\cdot \cdot \cdot \text{(iii)} \\
\end{align}$
Now calculating the sum of first m + n terms we get
${{S}_{m+n}}=\dfrac{m+n}{2}[2a+(m+n-1)d]$
Putting the value of (m + n – 1)d from equation (iii) we get,
$\begin{align}
  & {{S}_{m+n}}=\dfrac{m+n}{2}[2a+-2a-2] \\
 & \Rightarrow {{S}_{m+n}}=\dfrac{m+n}{2}(-2) \\
 & \Rightarrow {{S}_{m+n}}=-(m+n) \\
\end{align}$
So, we have proved that the sum of first m + n terms is - (m + n).

Note: Some students can make the mistake of adding equation (i) and (ii) to get the sum of m + n terms but this is wrong. It will not give the sum of m + n terms but it will give the sum of the sum of first n terms and sum of first m terms. That means some terms get added twice which is wrong.