If the sum of first p terms of an A.P is equal to the sum of first q terms, then find the sum of first (p + q) terms

Solution:

Let a and d be the first term and the common difference of the A.P. respectively.

We know that S_n = n/2 [2a + (n - 1) d]

Hence,

S_p = p/2 [2a + (p - 1) d]

S_q = q/2 [2a + (q - 1) d]

According to the question,

⇒ p/2 [2a + (p - 1) d] = q/2 [2a + (q - 1) d]

⇒ p [2a + ( p -1) d] = q [2a + (q -1) d ]

⇒ 2ap + pd ( p - 1) = 2aq + qd (q - 1)

⇒ 2a ( p - q) + d [p ( p - 1) - q (q - 1)] = 0

2a ( p - q) + d [p² - p - q² + q] = 0

⇒ 2a ( p - q) + d [(p - q)(p + q) - (p - q)] = 0

⇒ 2a ( p - q) + d [(p - q)(p + q - 1)] = 0

⇒ 2a + d (p + q - 1) = 0

d = - 2a/(p + q - 1) ....(1)

Now,

S_{p + q} = (p + q)/2 [2a + (p + q - 1) (- 2a/(p + q - 1)] [from (1)]

S_{p + q} = (p + q)/2 [2a - 2a]

= 0

Thus, the sum of the first (p + q) terms of the A.P is 0

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Question 10

If the sum of first p terms of an A.P is equal to the sum of first q terms, then find the sum of first (p + q) terms

Summary:

We are given that the sum of first p terms of an A.P is equal to the sum of first q terms, the sum of the first (p+q) terms is 0