Question

If sum of first m terms of ap be n and sum of first n terms be m then show that the sum of its first p+q terms is 0

Answer 1

Sum of first m terms = (m/2)[2a+(m-1)d]

Sum of its first n term= (n/2)[2a+(n-1)d]

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

2am +m²d-md = 2an+ n²d- nd

2a(m-n)+d(m²-n²)-d(m-n)=0

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

m -n = 0 Or [2a+(m+n-1)]=0

Sum of first (m+n) terms :-

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

= 0

Oh! Sorry I answered using m,n you can use p,q instead