Question

20. किसी A.P. के प्रथम p पदों के योग उसके प्रथम 4 पदों के
योग के बराबर है। तब उसके (pta) पदों का योग होगा।​

Answer 1

Correction: It will be q in place of 4 and we have to find the sum of (p + q) terms.

Solution:

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

Then the sum of the first p terms of the A.P. is

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

and the sum of the first q terms of the A.P. is

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

According to the question,

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

or, 2ap + p (p - 1)d = 2aq + q (q - 1)d

or, 2ap - 2aq = {q (q - 1) - p (p - 1)}d

or, 2a (p - q) = (q² - q - p² + p)d

or, 2a (p - q) = {- (p² - q²) + (p - q)}d

or, 2a (p - q) = - (p - q) (p + q - 1)d

or, 2a = - (p + q - 1)d, since p - q ≠ 0 ..... (1)

Now the sum of (p + q) terms is

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

= (p + q)/2 * [- (p + q - 1)d + (p + q - 1)d], by (1)

= (p + q)f2 * 0

= 0