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A sequence of numbers is written in the following fashion: 1, 7, 1, 1, 7, 7, 1, 1, 1, 7, 7, 7, ... till ‘n’ terms. The sum of first 5001 terms of the given sequence is,
a) 5001 b) 20478 c) 19911 d) 39822
Answer:
n(n+1) = 5000, n = 70,
total = 1/2(70*71 + 7*70*71) + 31.
70 terms of 7 and of 1 plus 31 more terms
35*71*(7+1) =19880 + 31
= 19911
Sum has to be odd so A or C
But obviously sum > 5001
Hence C.
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