Question

answer this question ​

Answer 1

Hope this helps u....

Answer 2

Answer:

Hope this helps.

First observe that the claim is not even true if m = n, for then the condition that the sum of the first m terms be equal to the sum of the first n terms is certainly satisfied for any A.P., but it is generally not the case that sum of the first m+n = 2n terms is zero.

Suppose then that m ≠ n.

Let a be the first term and let d be the common difference.

The sum of the first m terms is:  ma + m(m-1)d/2

The sum of the first n terms is:  na + n(n-1)d/2

That these are equal

=> ma + m(m-1)d/2 = na + n(n-1)d/2

=> (m-n)a + [ m(m-1) - n(n-1) ] d/2 = 0

=> (m-n)a + [ m²-n² -m +n ] d/2 = 0

=> (m-n)a + [ (m-n)(m+n) - (m-n) ] d/2 = 0

=> (m-n) [ a + (m+n-1)d/2 ] = 0

=> a + (m+n-1)d/2 = 0             [ can divide by m-n since m ≠ n ]

=> (m+n)a + (m+n)(m+n-1)d/2 = 0

=> sum of first m+n terms is zero.