Question 1003636
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1) Which of the following sequences are arithmetic?
Select one or more:
A. 1, -2, 4, -8, 16,-32,...
B. 20, 12, 4, -4, -12, ...
C. 1, 2, 4, 7, 11, ...
D. -10, -5, 1, 8, 16,...
E. 9.3, 9.9, 10.5, 11.1, 11.7,..
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B)  and  E)
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2)  How many terms of the sequence  2, 4, 6, 8, 10,. . .  yield a sum of  132?


It is the arithmetic progression with the first term  2  and the common difference  2. 

The formula for the sum of  n  first terms of an arithmetic progression is


{{{S[n]}}} = {{{(a[1] + ((n-1)*d)/2)*n}}}


(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Sequences-and-series/Arithmetic-progressions.lesson>Arithmetic progressions</A>&nbsp; in this site). 


In this case &nbsp;{{{S[n]}}} = {{{(2 + ((n-1)*2)/2)*n}}} = {{{2n + n*(n-1)}}}.


Therefore, &nbsp;to find &nbsp;n, &nbsp;we need to solve an equation 


2n + n*(n-1) = 132.


Simplify and solve:


{{{2n}}} + {{{n^2}}} - {{{n}}} = {{{132}}},


{{{n^2}}} +{{{n}}} - {{{132}}} = {{{0}}},


Factor:


(n-11)*(n+12) = 0,


The roots are &nbsp;n = 11 &nbsp;and &nbsp;n = -12.


Only &nbsp;n = 11 &nbsp;does suit.


<U>Answer</U>. &nbsp;n = 11.