Given that an arithmetic series has first term 51 and S_6=S_12. How do you find d and the maximum value of S_n?

1 Answer
Jan 7, 2017

d=-6 and maximum is S_9 = 243

Explanation:

The general term of an arithmetic sequence can be written:

a_n = a+d(n-1)

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:

S_n = n((a_1 + a_n)/2) = n ((a+(a+d(n-1)))/2) = na + 1/2dn(n-1)

In our example:

a = 51

0 = S_12-S_6 = (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 S_9:

S_9 = na + 1/2dn(n-1) = 9(51)+1/2(-6)(9)(9-1) = 243