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Difference between revisions of "2013 AMC 12A Problems/Problem 7"

(Created page with "<math>S_9 = 110</math>, <math>S_7 = 42</math> <math>S_8 = S_9 - S_ 7 = 110 - 42 = 68</math> <math>S_6 = S_8 - S_7 = 68 - 42 = 26</math> <math>S_5 = S_7 - S_6 = 42 - 26 = 16</m...")
 
(→‎Video Solution (private))
 
(4 intermediate revisions by 4 users not shown)
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== Problem==
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The sequence <math>S_1, S_2, S_3, \cdots, S_{10}</math> has the property that every term beginning with the third is the sum of the previous two.  That is, <cmath> S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3. </cmath> Suppose that <math>S_9 = 110</math> and <math>S_7 = 42</math>.  What is <math>S_4</math>?
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<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16\qquad </math>
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==Solution==
 
<math>S_9 = 110</math>, <math>S_7 = 42</math>
 
<math>S_9 = 110</math>, <math>S_7 = 42</math>
  
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<math>S_4 = S_6 - S_5 = 26 - 16 = 10</math>
 
<math>S_4 = S_6 - S_5 = 26 - 16 = 10</math>
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Therefore, the answer is <math>\boxed{\textbf{(C) }{10}}</math>
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==Video Solution==
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https://youtu.be/CCjcMVtkVaQ
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~sugar_rush
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 10:34, 25 September 2024

Problem

The sequence $S_1, S_2, S_3, \cdots, S_{10}$ 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 $S_4$?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16\qquad$

Solution

$S_9 = 110$, $S_7 = 42$

$S_8 = S_9 - S_ 7 = 110 - 42 = 68$

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

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

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

Therefore, the answer is $\boxed{\textbf{(C) }{10}}$

Video Solution

https://youtu.be/CCjcMVtkVaQ ~sugar_rush

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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