If demotes the sum of the first n terms of an A.P., prove that S30= 3 (S20 – S10) .
Proof: S20 = [2a + 19d]
= 10 [2a + 19d] (i)
S10 = [2a + 9d]
= 5 [2a + 9d] (ii)
L.H.S: S30 = [2a + 29d]
= 15 [2a + 29d]
R.H.S: 3 (S20 – S10)
= 3 [10 (2a + 19d) – 5 (2a + 9d)] {From (i) and (ii)}
= 15 [4a + 38d – 2a – 9d]
= 15 [2a + 29d]
Since, L.H.S = R.H.S
Hence, proved