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Re: The nth term (tn) of a certain sequence is defined as tn = t [#permalink]
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jezzsk8 wrote:
I think it should be \(t_n\,=\,t_{n-1}\,+\,4\).

In that case, it's given that \(t_1 = -7\). The \(0th\) term would be \(-11\), and the sequence could be expressed as..

\(t_n = -11 + 4n\)

The 71th term is...

\(t_{71} = -11 + 4(71)\)

\(t_{71} = 273\).

The answer is A.



Yes right. thanks for that, i have edited the main question.
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Re: The nth term (tn) of a certain sequence is defined as tn = t [#permalink]
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Thank you guys. Kudos

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Re: The nth term (tn) of a certain sequence is defined as tn = t [#permalink]
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The \(n\)th term (\(t_n\)) of a certain sequence is defined as \(t_n = t_{n-1} + 4\) . If \(t_1=-7\) then \(t_{71} =\)

A. 273

B. 277

C. 281

D. 283

E. 287
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Re: The nth term (tn) of a certain sequence is defined as tn = t [#permalink]
Carcass wrote:
The \(n\)th term (\(t_n\)) 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



t71 = -7 + 4(71-1) = -7 + 280 = 273

Answer choice A
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Re: The nth term (tn) of a certain sequence is defined as tn = t [#permalink]
Bunuel wrote:
The \(n_{th}\) term (\(t_n\)) 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

Kudos for correct solution.

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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Re: The nth term (tn) of a certain sequence is defined as tn = t [#permalink]
Carcass wrote:
The \(n\)th term (\(t_n\)) 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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Re: The nth term (tn) of a certain sequence is defined as tn = t [#permalink]
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