Problem:
Let a1​,a2​,…,ak​ be a finite arithmetic sequence with
a4​+a7​+a10​=17 and a4​+a5​+a6​+a7​+a8​+a9​+a10​+a11​+a12​+a13​+a14​=77.​
If ak​=13, then k=
Answer Choices:
A. 16
B. 18
C. 20
D. 22
E. 24
Solution:
In an arithmetic sequence with an odd number of terms, the middle term is the average of the terms. Since a4​,a7​,a10​ form an arithmetic sequence of three terms with sum 17,a7​=317​. Since a4​,a5​,…,a14​ form an arithmetic sequence of 11 terms whose sum is 77, the middle term, a9​=1177​=7. Let d be the common difference for the given arithmetic sequence. Since a7​ and a9​ differ by 2d,d=32​. Since a7​=a1​+6d, it follows that a1​=35​. From ak​=a1​+(k−1)d=35​+(k−1)32​=13 we obtain k=18.