Let Sn be the sum to n-terms of an arithmetic progression 3,7,11, . If 40< ( 6n(n+1) k=1n Sk )<42, then n equals .
9
Sk= k2[6+4(k-1)]=2k2+k
k=1nSk= 2n(n+1)(2n+1)6+ n(n+1)2= n(n+1)(4n+5)6
6n(n+1) k=1nSk=4n+5
40<4n+5<42 8.75<n<9.25 n=9
Answer: 9
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