2003 AIME II Problem 8

Question

Find the eighth term of the sequence

1440,

1716,

1848,

\ldots,

whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.

Answer

The product of two linear functions of

n

is a quadratic function of

n

Fit

an^2 + bn + c

through the three given terms at

n = 0,

1,

2,

then evaluate at

n = 7

The

n

th term of an arithmetic sequence is linear in

n,

so the product of corresponding terms of two arithmetic sequences is a quadratic

t_n = an^2 + bn + c.

Indexing the given terms by

n = 0,

1,

2:

\begin{aligned} c &= 1440, \\ a + b + c &= 1716, \\ 4a + 2b + c &= 1848, \end{aligned}

which give

a + b = 276

and

2a + b = 204,

so

a = -72,

b = 348,

c = 1440.

The eighth term is

t_7 = -72 \cdot 49

+ 348 \cdot 7

+ 1440 = 348.

(Indeed

t_n = (180 - 24n)(8 + 3n),

a product of two arithmetic sequences matching the given terms.)

Problem 8 in other years