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
Explanation:
The general term of an arithmetic sequence can be written:
a_n = a+d(n-1)an=a+d(n−1)
where
The sum of the first
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(n−1))2)=na+12dn(n−1)
In our example:
a = 51a=51
0 = S_12-S_6 = (12a+66d)-(6a+15d) = 6a+51d = 51(6+d)0=S12−S6=(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_9 = na + 1/2dn(n-1) = 9(51)+1/2(-6)(9)(9-1) = 243S9=na+12dn(n−1)=9(51)+12(−6)(9)(9−1)=243