Question

If demotes the sum of the first n terms of an A.P., prove that S₃₀= 3 (S₂₀ – S₁₀) .

Answer 1

Proof:

= 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