If the sum of the first p terms of an AP is the same as the sum of its first q terms (where p≠q) then show that the sum of its first (p+q) terms is zero.
Answer:
- We know that the sum of first n terms of an AP is given by Sn=n2(2a+(n−1)d), where a is the first term and n is the number of terms in the AP.
- We are given that Sp=Sq⟹p2(2a+(p−1)d)=q2(2a+(q−1)d)⟹p(2a+(p−1)d)=q(2a+(q−1)d)⟹2ap+(p−1)dp=2aq+(q−1)dq⟹2ap−2aq=(q−1)dq−(p−1)dp⟹2a(p−q)=q2d−dq−p2d+dp⟹2a(p−q)=q2d−p2d+dp−dq⟹2a(p−q)=−d(p2−q2)+d(p−q)⟹2a(p−q)=−d(p−q)(p+q)+d(p−q)⟹2a(p−q)=(p−q)[−d(p+q)+d]⟹2a=−d(p+q)+d⟹2a=(1−p−q)d…(i)
- Now, the sum of first (p+q) terms of the given AP is Sp+q=p+q2(2a+(p+q−1)d)=p+q2((1−p−q)d+(p+q−1)d) [Using(i)] =p+q2(d−pd−qd+pd+qd−d)=p+q2(0)=0
- Hence, the sum of (p+q) terms is 0 .