1.

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

Answer»

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



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