Sum to first l, m, n terms of A.P. are p, q, r. Prove that

Let the first term of AP be a and common difference be d.

Sum of the first p terms is:

S_l = × (2a + (l – 1)d) = p ………(1)

Sum of the first m terms is:

⇒ S_m = × (2a + (m – 1)d) = q ……… (2)

Sum of the first n terms is:

⇒ S_n = × (2a + (n – 1)d) = r ……… (3)

Now, (1) × + (2) × + (3) ×

We get, p × + q × + r ×

= 0 + [(m – n)l + (n – l)m + (l – m)n –m + n – n + l – l + m]

= [lm – ln + mn – lm + ln – mn + 0]

= × 0 = 0

∴ it is proved that, p × + q × + r × = 0