Question

- + +0. Inu u Tuu If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + 9) terms. A.P. are a, and e, respectively​

Answer 1

Answer:

Correct answer is 0

Step-by-step explanation:

Sp = Sq

⇒ p/2 (2a+(p−1)d) = q/2(2a+(q−1)d)

⇒ p/2 (2a+(p−1)d) = q/2(2a+(q−1)d)⇒ p(2a+(p−1)d) = q(2a+(q−1)d)

⇒ p/2 (2a+(p−1)d) = q/2(2a+(q−1)d)⇒ p(2a+(p−1)d) = q(2a+(q−1)d)⇒ 2ap+p^2d−pd = 2aq+q^2d−qd

⇒ p/2 (2a+(p−1)d) = q/2(2a+(q−1)d)⇒ p(2a+(p−1)d) = q(2a+(q−1)d)⇒ 2ap+p^2d−pd = 2aq+q^2d−qd⇒ 2a(p−q)+(p+q)(p−q)d−d(p−q) = 0

⇒ p/2 (2a+(p−1)d) = q/2(2a+(q−1)d)⇒ p(2a+(p−1)d) = q(2a+(q−1)d)⇒ 2ap+p^2d−pd = 2aq+q^2d−qd⇒ 2a(p−q)+(p+q)(p−q)d−d(p−q) = 0⇒ (p−q)[2a+(p+q)d−d] = 0

⇒ p/2 (2a+(p−1)d) = q/2(2a+(q−1)d)⇒ p(2a+(p−1)d) = q(2a+(q−1)d)⇒ 2ap+p^2d−pd = 2aq+q^2d−qd⇒ 2a(p−q)+(p+q)(p−q)d−d(p−q) = 0⇒ (p−q)[2a+(p+q)d−d] = 0⇒ 2a+(p+q)d−d = 0

⇒ p/2 (2a+(p−1)d) = q/2(2a+(q−1)d)⇒ p(2a+(p−1)d) = q(2a+(q−1)d)⇒ 2ap+p^2d−pd = 2aq+q^2d−qd⇒ 2a(p−q)+(p+q)(p−q)d−d(p−q) = 0⇒ (p−q)[2a+(p+q)d−d] = 0⇒ 2a+(p+q)d−d = 0⇒ 2a+((p+q)−1)d = 0

⇒ p/2 (2a+(p−1)d) = q/2(2a+(q−1)d)⇒ p(2a+(p−1)d) = q(2a+(q−1)d)⇒ 2ap+p^2d−pd = 2aq+q^2d−qd⇒ 2a(p−q)+(p+q)(p−q)d−d(p−q) = 0⇒ (p−q)[2a+(p+q)d−d] = 0⇒ 2a+(p+q)d−d = 0⇒ 2a+((p+q)−1)d = 0⇒ Sp+q = 0