The sequence a_1, a_2, a_3, … satisfies a_1 = 19, a_9 = 99, and, for all n ≥ 3, a_n is the arithmetic mean of the first n - 1 terms. Find a_2.

1999 AMC 12 Problem 20

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20.

The sequence

a_1,

a_2,

a_3,

\ldots

satisfies

a_1 = 19,

a_9 = 99,

and, for all

n \ge 3,

a_n

is the arithmetic mean of the first

n - 1

terms. Find

a_2.

29

59

79

99

179

Show that

a_n

is constant for all

n \ge 3

Then

a_3 = a_9 = 99

and

a_3 = \dfrac{a_1 + a_2}{2}

For

n \ge 3,

(n - 1)a_n = a_1 + \cdots + a_{n-1}.

Then

a_{n+1} = \dfrac{(n-1)a_n + a_n}{n} = a_n,

so the sequence is constant from

a_3

onward. Hence

a_3 = a_9 = 99.

Since

a_3 = \dfrac{a_1 + a_2}{2} = \dfrac{19 + a_2}{2} = 99,

we get

a_2 = 179.

Thus, the correct answer is E.