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The nth term (tn) of a certain sequence is defined as tn = t(n-1) + 4

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shrive555
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The nth term (tn) of a certain sequence is defined as tn = t(n-1) + 4 [#permalink] New post   Updated on: 22 Oct 2018, 01:22
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The nth term (tn) of a certain sequence is defined as \(t_n = t_{n-1} + 4\). If \(t_1= -7\) then \(t_{71} =\)

A. 273
B. 277
C. 281
D. 283
E. 287
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A

Originally posted by shrive555 on 24 Nov 2010, 10:50.
Last edited by Bunuel on 22 Oct 2018, 01:22, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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BrentGMATPrepNow
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Re: The nth term (tn) of a certain sequence is defined as tn = t(n-1) + 4 [#permalink] New post   03 Mar 2020, 07:20
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shrive555 wrote:
The nth term (tn) of a certain sequence is defined as \(t_n = t_{n-1} + 4\). If \(t_1= -7\) then \(t_{71} =\)

A. 273
B. 277
C. 281
D. 283
E. 287


If we don't already know the formula for arithmetic sequences, we can still answer the question.

The given formula tells us that each term in the sequence is 4 greater than the term before it.
So, let's list some terms to see if we can see a pattern

term1 = -7
term2 = -7 + 4
term3 = -7 + 4 + 4
term4 = -7 + 4 + 4 + 4
term5 = -7 + 4 + 4 + 4 + 4
.
.
.
At this point we can probably see the pattern

So, term71 = -7 + 4 + 4 + 4 + 4 + 4 + 4 + 4.......

QUESTION: How many 4's are in the sum for term71?
Well, term1 has 0 4's
term2 has 1 4
term3 has 2 4's
term4 has 3 4's
etc

So, we can see that term71 has 70 4's

In other words, term71 = -7 + (70)(4)
= -7 + 280
= 273

Answer: A

Cheers,
Brent
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Bunuel
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The nth term (tn) of a certain sequence is defined as tn = t(n-1) + 4 [#permalink] New post   24 Nov 2010, 10:59
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shrive555 wrote:
The nth term (tn) of a certain sequence is defined as
tn = tn-1 + 4. If t1= -7 then t71 =

OA=273


Please post the whole question with answer choices.

Given:

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

Then:

\(t_2=t_1+4\);

\(t_3=t_2+4=t_1+2*4\);

\(t_4=t_3+4=t_1+3*4\);

...

\(t_n=t_1+(n-1)*4\).

So, \(t_{71}=t_1+70*4=-7+280=273\).
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Wayxi
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The nth term (tn) of a certain sequence is defined as tn = t(n-1) + 4 [#permalink] New post   24 Nov 2010, 19:51
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Here is another way:

Given: T1 = Tn-1 +4 & T1 = -7

Then solve to get the
Direct Formula:
Tn = k*n+x k is the constant change which is 4 in this problem, find x

T1 = k + x
T1: -7 = 4 + x
x = -11

Tn= 4*n -11

Use this formula to find T71:
T71= 4*(71) -11
= 284 - 11
= 273
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Re: The nth term (tn) of a certain sequence is defined as tn = t(n-1) + 4 [#permalink] New post   24 Nov 2010, 19:59
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shrive555 wrote:
The nth term (tn) of a certain sequence is defined as
tn = tn-1 + 4. If t1= -7 then t71 =


OA=273


I see you are working with sequences and series today. The story I wrote in response to your previous question will work perfectly well here too since this is an arithmetic progression too. If you understand it, you will get this question on your own. Go back and read it again and try this question again. If you don't get it, let me know what you don't understand in the theory.
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Re: The nth term (tn) of a certain sequence is defined as tn = t(n-1) + 4 [#permalink] New post   04 Apr 2022, 02:57
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Re: The nth term (tn) of a certain sequence is defined as tn = t(n-1) + 4 [#permalink]
04 Apr 2022, 02:57
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