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Math Expert V
Joined: 02 Sep 2009
Posts: 58409
The nth term, t(n), of a certain sequence is defined as t(n) = t(n-1)  [#permalink]

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Difficulty:   5% (low)

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

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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

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Manager  G
Joined: 05 Feb 2016
Posts: 168
Location: India
Concentration: General Management, Marketing
WE: Information Technology (Computer Software)
Re: The nth term, t(n), of a certain sequence is defined as t(n) = t(n-1)  [#permalink]

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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
Senior SC Moderator V
Joined: 22 May 2016
Posts: 3565
The nth term, t(n), of a certain sequence is defined as t(n) = t(n-1)  [#permalink]

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

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Instructions for living a life. Pay attention. Be astonished. Tell about it. -- Mary Oliver 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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