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.