Question

In A.P. If $ {{S}_{m}}=n $ and $ {{S}_{n}}=m $ find $ {{S}_{m+n}} $ .

Answer 1

Hint:Arithmetic Progression is a sequence of numbers such that the difference between consecutive terms is the same (constant). It is also called arithmetic sequence. It has a form of following-
a, a+d, a+2d, a+3d………………
Where ‘a’ is the first term and d is the difference which is always constant. In a AP, sum of first n term is- \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]\] .We will use this formula to find required solution.

Complete step-by-step answer:
It is given that,
Sum of first n terms is m.
Apply here above formula,
 \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]=m\] …..(1)
And, the sum of the first m terms is n.
\[{{S}_{m}}=\dfrac{m}{2}\left[ 2a+(m-1)d \right]=n\] ….(2)

We need to find the sum of first m+n terms.
 $ {{S}_{m+n}}=? $
Sum of first m+n terms is-
 $ {{S}_{m+n}}=\dfrac{m+n}{2}\left[ 2a+(m+n-1) \right] $ ….(3)

Subtracting (1) from (2),
 $ {{S}_{n}}-{{S}_{m}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]-\dfrac{m}{2}\left[ 2a+(m-1)d \right] $
 $ \begin{align}
  & 2(m-n)=2a(n-m)+[({{n}^{2}}-{{m}^{2}})-(n-m)]d \\
 & -2(n-m)=(n-m)[2a+\{(n+m)-1\}d] \\
\end{align} $
 Divide (n-m) to both sides.
 $ -2=2a+\{(n+m)-1\}d $
To make LHS like equation 3 LHS we need to multiply it by (n+m) then divide by 2.
 $ -(m+n)=\dfrac{m+n}{2}[2a+\{(n+m)-1\}d] $
 $ {{S}_{m+n}}=-(m+n) $

The sum of the m+n term is -(m+n).

Note:- We can also start by using sum of (m+n)the terms formula then try to split it in sum of nth terms and sum of mth terms formula by adding and subtracting some known variable. We should take care of substation of variables.