Difference between revisions of "2003 AIME II Problems/Problem 8"

(Added an alternate Solution)
(Solution)
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<math>9a+3b+c=1848</math>
 
<math>9a+3b+c=1848</math>
  
Solving gives a=-72, b=492, and c=1020. Thus, the answer is <math>-72(8)^2+492\cdot8+1020=348</math>
+
Solving gives <math>a=-72, b=492,</math> and <math>c=1020</math>. Thus, the answer is <math>-72(8)^2+492\cdot8+1020=348</math>.
  
 
==Alternate Solution==
 
==Alternate Solution==

Revision as of 00:58, 27 February 2011

Problem

Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.

Solution

If you multiply the corresponding terms of two arithmetic sequences, you get the terms of a quadratic function. Thus, we have a quadratic $ax^2+bx+c$ such that $f(1)=1440$, $f(2)=1716$, and $f(3)=1848$. Plugging in the values for x gives us a system of three equations:

$a+b+c=1440$

$4a+2b+c=1716$

$9a+3b+c=1848$

Solving gives $a=-72, b=492,$ and $c=1020$. Thus, the answer is $-72(8)^2+492\cdot8+1020=348$.

Alternate Solution

Setting one of the sequences as $a+nr_1$ and the other as $b+nr_2$, we can set up the following equalities

$ab = 1440$

$(a+r_1)(b+r_2)=1716$

$(a+2r_1)(b+2r_2)=1848$

We want to find $(a+7r_1)(b+7r_2)$

Foiling out the two above, we have

$ab + ar_2 + br_1 + r_1r_2 = 1716$ and $ab + 2ar_2 + 2br_1 + 4r_1r_2 = 1848$

Plugging in $ab=1440$ and bringing the constant over yields

$ar_2 + br_1 + r_1r_2 = 276$

$ar_2 + br_1 + 2r_1r_2 = 204$

Subtracting the two yields $r_1r_2 = -72$ and plugging that back in yields $ar_2 + br_1 = 348$

Now we find

$(a+7r_1)(b+7r_2) = ab + 7(ar_2 + br_1) + 49r_1r_2 = 1440 + 7(348) + 49(-72) = \boxed{348}$

See also

2003 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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