Q73 of 84 Page 5

If demotes the sum of the first n terms of an A.P., prove that S30= 3 (S20 – S10). (CBSE 2014)

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

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