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3. Find the minimum value of n such that the sum of the first n terms of a G.P. 1, 2, 22, 23, ... .is greater than or equal to 2000.у уChow0 in 121 |
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Answer» n=6Sn = a(R^n-1) / r-1 (as r>1)here a=1 r=3 and Sn < 365 => 1{(3^n)-1} / 3-1 < 365 => (3^n)-1 < 730 => 3^n < 731Now, 3^6 = 729 which is less than 731, THEREFORE n=6 |
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