Let S_n be the sum to n-terms of an arithmetic progression 3 ,7 ,11 , … … If 40

Question

Let S_n be the sum to n-terms of an arithmetic progression 3 ,7 ,11 , … … If 40< ( 6n (n+1 ) _k=1^n 0.05ems_k )<42, then n equals

Accepted Answer

Correct answer: 9. (9) S_n=3+7+11+ … …n terms= n2 (6+ (n-1 )4 )=3n+2n^2-2n=2n^2+n _k=1^n 0.05emS_k=2 _k=1^n 0.05emK^2+ _k=1^nK=2 n (n+1 ) (2n+1 )6+ n (n+1 )2=n (n+1 )( 2n+13+ 12)= n (n+1 ) (4n+5 )6 ⇒40< 6n (n+1 ) _k=1^n 0.05emS_k<4240<4n+5<4235<4n<37n=9

$Let S_{n} be the sum to n-terms of an arithmetic progression 3, 7, 11, \dots \dots If 40 < (\frac{6}{n ( n + 1 )} \sum_{k = 1}^{n} s_{k}) < 42, then n equals$

Solution

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