Question

If demotes the sum of first n terms of an A.P., prove that S₁₂ = 3 (s₈ – S₄) .

Answer 1

S₈ = \(\frac{8}{2}\)[2a + 7d] 

= 4 (2a + 7d).....  (i) 

S₄ = \(\frac{4}{2}\)[2a + 7d] 

= 2 (2a + 7d)....... (ii) 

L.H.S = S₁₂ = \(\frac{12}{2}\)[2a + 11d] 

= 6 [2a + 11d] 

R.H.S = 3 (S₈ – S₄) 

= 3 [4 (2a + 7d) – 2 (2a + 3d)] 

[From (i) and (ii)]

= 6 [4a + 14d – 2a – 3d] 

= 6 [2a + 11d] 

Since, L.H.S = R.H.S 

Hence, proved