Given that an arithmetic series has first term 51 and S6=S12. How do you find d and the maximum value of Sn?

1 Answer
Jan 7, 2017

d=6 and maximum is S9=243

Explanation:

The general term of an arithmetic sequence can be written:

an=a+d(n1)

where a is the initial term and d the common difference.

The sum of the first n terms is the average term multiplied by the number of terms. The average term is the same as the average of the first and last terms.

So we find:

Sn=n(a1+an2)=n(a+(a+d(n1))2)=na+12dn(n1)

In our example:

a=51

0=S12S6=(12a+66d)(6a+15d)=6a+51d=51(6+d)

So:

d=6

The sequence starts:

51,45,39,33,27,21,15,9,3,3,9,15

The maximum value of the sum is S9:

S9=na+12dn(n1)=9(51)+12(6)(9)(91)=243