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If the number of ways of selecting 3 numbers out of `1, 2, 3, ……., 2n+1` such that they are in arithmetic progression in 441, then they are in arithemtric progression is 441, then the sum of the divisors of n is equal toA. 21B. 22C. 32D. None of these |
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Answer» Correct Answer - 3 `C_(2) +^(n+1)C_(2)=441 rArr (n(n-1))/(2)+((n+1)n)/(2)=441` `n^(2)=441 rArr n =21` `"sum of divisors of" 21 = (1+3)(1+7)` `=4xx8=32` |
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