Q64 of 202 Page 9

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

S8 = [2a + 7d]

= 4 (2a + 7d) (i)


S4 = [2a + 7d]


= 2 (2a + 7d) (ii)


L.H.S = S12 = [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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