Given that an arithmetic series has first term 51 and S_6=S_12S6=S12. How do you find d and the maximum value of S_nSn?

1 Answer
Jan 7, 2017

d=-6d=6 and maximum is S_9 = 243S9=243

Explanation:

The general term of an arithmetic sequence can be written:

a_n = a+d(n-1)an=a+d(n1)

where aa is the initial term and dd the common difference.

The sum of the first nn 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)Sn=n(a1+an2)=n(a+(a+d(n1))2)=na+12dn(n1)

In our example:

a = 51a=51

0 = S_12-S_6 = (12a+66d)-(6a+15d) = 6a+51d = 51(6+d)0=S12S6=(12a+66d)(6a+15d)=6a+51d=51(6+d)

So:

d=-6d=6

The sequence starts:

51, 45, 39, 33, 27, 21, 15, 9, 3, -3, -9, -1551,45,39,33,27,21,15,9,3,3,9,15

The maximum value of the sum is S_9S9:

S_9 = na + 1/2dn(n-1) = 9(51)+1/2(-6)(9)(9-1) = 243S9=na+12dn(n1)=9(51)+12(6)(9)(91)=243