Let a_1, a_2, …, a_k be a finite arithmetic sequence with a_4+a_7+a_10=17 and &a_4+a_5+a_6+…+a_12 & +a_13+a_14=77. If a_k=13, then k=

1993 AMC 12 Problem 21

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21.

Let

a_1,

a_2,

\ldots,

a_k

be a finite arithmetic sequence with

a_4+a_7+a_{10}=17

and

\begin{aligned} &a_4+a_5+a_6+\cdots+a_{12}\\ &\quad{}+a_{13}+a_{14}=77. \end{aligned}

a_k=13,

then

k=

16

18

20

22

24

Use the middle terms to find

a_7

and

a_9

The difference between

a_7

and

a_9

determines the common difference

Because symmetric terms of an arithmetic sequence average to the middle term,

3a_7=17,\qquad 11a_9=77.

Thus

a_7=\frac{17}{3}

and

a_9=7,

so the common difference is

d=\frac{7-\frac{17}{3}}{2}=\frac{2}{3}.

Since

13-7=6=9d,

we have

a_k=a_{9+9}=a_{18}.

Thus the correct answer is B.