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In a certain sequence, each term is 7 greater than the term before it.

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In a certain sequence, each term is 7 greater than the term before it.  [#permalink]

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New post 21 Jan 2019, 02:24
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In a certain sequence, each term is 7 greater than the term before it. If the third term of the sequence is 39, what is the 13th term?

A 102
B 109
C 116
D 123
E 130

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In a certain sequence, each term is 7 greater than the term before it.  [#permalink]

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New post 21 Jan 2019, 08:25
Bunuel wrote:
In a certain sequence, each term is 7 greater than the term before it. If the third term of the sequence is 39, what is the 13th term?

A 102
B 109
C 116
D 123
E 130



a1=x
a2=x+7
a3=x+14
.
.
.
a13=x+84

given a3= 39
39=x+14
x=25
a13= 25+84 = 109

or use formula
an=a+(n-1)*d
a13=25+ ( 13-1) * 7
a13= 25 + 84
a13= 109
IMO B
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In a certain sequence, each term is 7 greater than the term before it.  [#permalink]

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New post 21 Jan 2019, 08:42
Bunuel wrote:
In a certain sequence, each term is 7 greater than the term before it. If the third term of the sequence is 39, what is the 13th term?

A 102
B 109
C 116
D 123
E 130


Above part signifies that the series will be an AP with d = 7 and in AP => \(a_n = a + (n-1) d\)

a => First term , d => common difference

\(a_3\) will be =>\(a_3 = a + 2d\)
=> \(39 - 14 = a_3\)
=> \(a_3 = 25\)

\(a_{13}\) = 25 + 12*7 = 109

Answer B
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Re: In a certain sequence, each term is 7 greater than the term before it.  [#permalink]

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New post 29 Jan 2019, 17:30
13 - 3 = 10

10*7 =70

39 + 70 = 109
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Re: In a certain sequence, each term is 7 greater than the term before it.   [#permalink] 29 Jan 2019, 17:30
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