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Difference between revisions of "2006 iTest Problems/Problem 7"

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{{iTest box|year=2006|num-b=5|num-a=7|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
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{{iTest box|year=2006|num-b=6|num-a=8|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 22:29, 27 November 2018

Problem

The sum of $17$ consecutive integers is $2006$. Find the second largest integer.

$\mathrm{(A)}\,17\quad\mathrm{(B)}\,72\quad\mathrm{(C)}\,95\quad\mathrm{(D)}\,101\quad\mathrm{(E)}\,102\quad\mathrm{(F)}\,111\quad\mathrm{(G)}\,125$

Solution

Let $a$ be the first number in the series, so $a + (a + 1) + \cdots (a + 15) + (a+16) = 2006$. By combining like terms, we can solve for $a$. \begin{align*} 17a + (17 \cdot 8) &= 2006 \\ a + 8 &= 118 \\ a &= 110 \end{align*} Since the first term is $110$, the second to last term is $110+15 = \boxed{\textbf{(G) } 125}$.

See Also

2006 iTest (Problems, Answer Key)
Preceded by:
Problem 6 		Followed by:
Problem 8
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