Question

If the sum of m terms is n and the sum of n terms is m then show that sum of first m+ n terms is -(m+n) ​

Answer 1

Answer:

the answer will be -m+(-n)

Answer 2

Answer:

Sum of first (m + n) terms of AP is - (m+n). ​

Step-by-step explanation:

We know that Sum of n terms of AP, $S_n=\frac{n}{2} [2a+(n-1)d]$  where, a = First term and d = common difference.

We are given that sum of m terms is n and the sum of n terms is m.

Sum of m terms of an AP, $S_m$ = $\frac{m}{2} [2a+(m-1)d]$

                            n = $\frac{m}{2} [2a+(m-1)d]$

                           $2n = 2am +m(m-1)d$

                           $2n = 2am +(m^{2} -m)d$ ---------- [Equation 1]

Sum of n terms of an AP, $S_n$ = $\frac{n}{2} [2a+(n-1)d]$

                            m = $\frac{n}{2} [2a+(n-1)d]$

                           $2m = 2an +n(n-1)d$

                           $2m = 2an +(n^{2} -n)d$ ---------- [Equation 2]

Subtract equation 2 from 1, we get;

        $2n-2m = 2am +(m^{2} -m)d - 2an- (n^{2} -n)d$

        $2(n-m) = 2a(m-n) +((m^{2} -n^{2}) -(m-n))d$

        $-2(m-n) = 2a(m-n) +((m+n)(m-n) -(m-n))d$

Taking (m - n) common from right side of equation;

        $-2(m-n) = (m-n)(2a +((m+n) -1)d$

                $-2 = 2a +((m+n) -1)d$

Multiply both sides by $\frac{m+n}{2}$ we get,

                 $-2*\frac{(m+n)}{2} = \frac{m+n}{2}* [2a +((m+n) -1)d]$

                   $\frac{m+n}{2} [2a +((m+n) -1)d] = -(m+n)$

The left side of equation is the formula for sum of first (m+n) terms of AP which comes out to be -(m+n) .