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The nth term, t(n), of a certain sequence is defined as t(n) = t(n-1)

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The nth term, t(n), of a certain sequence is defined as t(n) = t(n-1)  [#permalink]

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New post 02 Apr 2018, 01:45
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Question Stats:

83% (01:20) correct 17% (02:42) wrong based on 75 sessions

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Re: The nth term, t(n), of a certain sequence is defined as t(n) = t(n-1)  [#permalink]

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New post 02 Apr 2018, 03:02
Bunuel wrote:
The \(n_{th}\) term, \(t_n\), of a certain sequence is defined as \(t_n = t_{n-1} + 4\). If \(t_1 = -11\), then \(t_{82} =\)

(A) 313
(B) 317
(C) 320
(D) 321
(E) 340



t2=t1+4
t3=t2+4
.
.
t82=t81+4

adding all we get t82=t1+81*4, since t1=-11
t82=313.

option A
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The nth term, t(n), of a certain sequence is defined as t(n) = t(n-1)  [#permalink]

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New post 02 Apr 2018, 19:42
Bunuel wrote:
The \(n_{th}\) term, \(t_n\), of a certain sequence is defined as \(t_n = t_{n-1} + 4\). If \(t_1 = -11\), then \(t_{82} =\)

(A) 313
(B) 317
(C) 320
(D) 321
(E) 340

This problem looks like a regular arithmetic progression.

The standard equation for AP is usually detectable in questions like this one.

\(t_n = t_{n-1} + 4\)

\(t_1 = -11\)
\(t_2 =(t_{1} + 4)=(-11+4)= -7\)
\(t_3 = t_{2} + 4 = -3\)
\(t_4 = t_{3} + 4 = 1\)

The above is equivalent to . . . (looking for standard arithmetic sequence)

\(t_1 = -11\)
\(t_2 = t_1 + 4 = -7\)
\(t_3 = t_1 + 4 + 4 = -3\)
\(t_4 = t_1 + 4 + 4 + 4 = -1\)

Extrapolate the standard equation of arithmetic sequence.
The number of 4s added is (n - 1)
4 = common difference, d

\(a_n = a_1 + (n-1)d\)

Rewrite:
\(t_n = t_1 + (n-1)4\)
\(t_1\) is -11

Hence
\(t_{82} = -11 + (81*4) =\)
\((-11 + 324) = 313\)

Answer A
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The nth term, t(n), of a certain sequence is defined as t(n) = t(n-1)   [#permalink] 02 Apr 2018, 19:42
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