If the sum of the first p terms of an AP be q and the sum of its first q terms be p then show that the sum of its first (p+q) terms is −(p+q).
Answer:
- Let a be the first term and d be the common difference of the given AP. Then, Sp=q⟹p2(2a+(p−1)d)=q⟹2ap+p(p−1)d=2q…(i) And, Sq=p⟹q2(2a+(q−1)d)=p⟹2aq+q(q−1)d=2p…(ii)
- On subtracting (ii) from (i), we get [2ap+p(p−1)d]−[2aq+q(q−1)d]=2q−2p⟹2a(p−q)+(p2−p−q2+q)d=2q−2p⟹2a(p−q)+(p2−q2)d−(p−q)d=−2(p−q)⟹2a(p−q)+(p−q)(p+q)d−(p−q)d=−2(p−q)⟹2a+(p+q)d−d=−2⟹2a+(p+q−1)d=−2…(iii)
- Now, the sum of the first (p+q) terms of the AP is Sp+q=p+q2(2a+(p+q−1)d)=p+q2(−2)[Using (iii)]=−(p+q)
- Thus, the sum of the first (p+q) terms is −(p+q).