2013 AMC 12A Problem 7

Attempt Problem 7 of the 2013 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

The sequence

S_1,

S_2,

S_3,

\ldots,

S_{10}

has the property that every term beginning with the third is the sum of the previous two. That is,

S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3.

Suppose that

S_9 = 110

and

S_7 = 42.

What is

S_4?

4

6

10

12

16

Rearrange the rule to

S_{n-2} = S_n - S_{n-1}

and step downward

From

S_9 = S_7 + S_8

find

S_8,

then work down to

S_4

Since

S_9 = S_7 + S_8,

we get

S_8 = 110 - 42 = 68.

Then

S_6 = S_8 - S_7 = 68 - 42 = 26,

S_5 = S_7 - S_6 = 42 - 26 = 16,

and

S_4 = S_6 - S_5 = 26 - 16 = 10.

Thus, the correct answer is C.

Problem 7 in other years