Question

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`?

Answer 1

`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`