Question

if in an AP the sum of m terms is equal to n terms and the sum of n terms is equal to m then prove that the sum of m + n terms is - ( m + n).​

Answer 1

Answer:

Let a be the first term and d be the common difference of the given A.P

Then,

$s_{m+n}$

$⇒\frac{m}{2} {2a+(m-1)d}=n$

$⇒2am + m(n-1)d=2n$ -------(i)

$and, s_{n}=m$

$⇒\frac{n}{2} {2a+(n-)d}$

$⇒2an +n(n-1)d=2m$-------(ii)

subtracting equation (ii) from equation (i),we get

$⇒2a(m-n)+{m(m-1)-n(n-1)}d = 2n-2m$

$⇒2a(m-n)+{(m²-n^{2})-n(n-1)}d= -2(m-n)$

$⇒2a+(m+n-1)d= -2$        [on dividing both sides by (m-n)]-------(iii)

Now,

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

$⇒s_{m+n}=\frac{m+n}{2} (-2)$           [using(iii)]

$⇒s_{m+n}= -(m+n)$

$HENCE PROVED$

Answer 2

Answer:

ANS

Step-by-step explanation:

Let a be the first term and d be the common difference of the given A.P

Then,

-------(i)

-------(ii)

subtracting equation (ii) from equation (i),we get

       [on dividing both sides by (m-n)]-------(iii)

Now,

          [using(iii)]