Question

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 ___________

Answer 1

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