Question

If the sum of p terms of an AP is the same as the sum of q terms, show that the sum of (p + q) terms is zero.

Answer 1

`S_n = n/2[2a + (n - 1)d]`

Set the sum of p terms equal to the sum of g terms

Given S_p = S_q

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

p[2a + (p − 1)d] = q[2a + (q − 1)d]

Expand both sides:

2ap + p(p − 1)d = 2aq + q(q −  1)d

2ap + (p² − p)d = 2aq + (q² − q)d

2a(p − q) + [p² − p (q² − q)]d = 0

2a(p − q) + [p² − q² − (p − q)]d = 0

p² − q² = (p − q)(p + q)

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

(p − q)[2a + (p + q − 1)d] = 0

Since p ≠ q (as they are distinct numbers of terms), we can divide by (p − q):

2a + (p + q − 1)d = 0    .....(1)

Using the sum formula for (p + q) terms:

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

Substitute the value from Equation 1 into this formula:

`S_(p + q) = (p + q)/2 xx [0]`

`S_(p + q) = 0`

Hence, the sum of its first (p + q) terms is zero.