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
Explanation:
The general term of an arithmetic sequence can be written:
an=a+d(n−1)
where
The sum of the first
So we find:
Sn=n(a1+an2)=n(a+(a+d(n−1))2)=na+12dn(n−1)
In our example:
a=51
0=S12−S6=(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=na+12dn(n−1)=9(51)+12(−6)(9)(9−1)=243