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2008 AMC 10B Problem 11

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2008 AMC 10B Problem 11AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 2 of 5 · about 2 min · position P11-15

Suppose that $(u_n)$ is a sequence of real numbers satisfying $u_{n+2}=2u_{n+1}+u_n$ ,

and that $u_3=9$ and $u_6=128$ . What is $u_5$ ?

A)
$40$
B)
$53$
C)
$68$
D)
$88$
E)
$104$

Answer

Key insight

Call u_4 = a; then u_5 = 2a + 9 and u_6 = 5a + 18, so 5a + 18 = 128 gives a = 22 and u_5 = 53.

Solution

The unknown term between the two given ones is $u_4$ ; call it $a$ . Apply the recurrence twice:
$u_5 = 2u_4 + u_3 = 2a + 9,$
$u_6 = 2u_5 + u_4 = 2(2a + 9) + a = 5a + 18.$
Setting $u_6 = 128$ gives $5a + 18 = 128$ , so $a = 22$ . Then
$u_5 = 2(22) + 9 = 53.$
Check: $u_6 = 2(53) + 22 = 128$ .

The answer is $\boxed{\textbf{(B)}\ 53}$ .

Why this works

A second-order recurrence is determined by any two consecutive terms, but here the givens are three apart. Naming the missing term as a variable and expressing later terms in it converts the recurrence into a single linear equation. Substituting a symbol for the gap, rather than guessing, is the general move.

Alternative approach

Work backwards symbolically: $u_4 = u_6 - 2u_5$ , and $u_5 = 2u_4 + u_3 = 2(128 - 2u_5) + 9$ , so $5u_5 = 265$ and $u_5 = 53$ .

The trap

Trying to run the recursion backwards from u_6 without a second known term, or mislabeling which terms 9 and 128 are.

Common mistakes

Trying to run the recursion backwards from u_6 without a second known term, or mislabeling which terms 9 and 128 are.
Writing the recurrence as $u_{n+2} = 2u_n + u_{n+1}$ (swapping the coefficients), which gives $u_5 = 40$ .

Techniques

Set up the equation/formula and compute; no special trick needed