The first term of an AP is 15, and the common difference is 6. The smallest value of , such that the sum to terms of this AP exceeds 2016 is ___________
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Step-by-step explanation:
a=15 d=6 n=?
n/2{2a+(n-1)d}=2016
n/2{2*15+(n-1)6}=2016
n/2{30+6n-6}=2016
n/2{24+6n}=2016
6n^2+24n=4032
n^2+4n-672=0
n^2+36n-32n-672=0
n(n+36)-32(n+36)=0
(n+36)(n-32)
n-32=0
n=32
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