If the sum of the first pp terms of an AP be qq and the sum of its first qq terms be pp then show that the sum of its first (p+q)(p+q) terms is −(p+q)−(p+q).
Answer:
- Let aa be the first term and dd 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).