Let a1, a2, a3.....a11 be real numbers satisfying
a1 = 15, 27 - 2a2 > 0 and ak = 2ak_1 – ak–2 for k = 3, 4..... 11.
If , then the value of is equal to
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ak – ak – 1 = ak – 1 – ak – 1 an's are in A.P.
Let common difference d
on solving d = – 3, – 9 (rejected)
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