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In a sequence of 40 numbers, each term, except for the first one, is 7

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In a sequence of 40 numbers, each term, except for the first one, is 7 [#permalink]

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New post 01 Aug 2017, 13:27
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In a sequence of 40 numbers, each term, except for the first one, is 7 less than the previous term. If the greatest term in the sequence is 281, what is the smallest term in the sequence?

A. 8
B. 7
C. 1
D. 0
E. −6
[Reveal] Spoiler: OA

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Re: In a sequence of 40 numbers, each term, except for the first one, is 7 [#permalink]

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New post 01 Aug 2017, 13:39
i think it should be A

greatest term is 281 which would also be the first term

now we have to find last term that is least among all

difference between first and last term=

39*7=273

so last term is 281-273= 8
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Re: In a sequence of 40 numbers, each term, except for the first one, is 7 [#permalink]

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New post 01 Aug 2017, 14:17
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Bunuel wrote:
In a sequence of 40 numbers, each term, except for the first one, is 7 less than the previous term. If the greatest term in the sequence is 281, what is the smallest term in the sequence?

A. 8
B. 7
C. 1
D. 0
E. −6


281+(40-1)(-7)=8
A

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Re: In a sequence of 40 numbers, each term, except for the first one, is 7 [#permalink]

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New post 01 Aug 2017, 15:37
281-(7*39) answer A , think if it have 2 terms then calculate for 40 terms

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Re: In a sequence of 40 numbers, each term, except for the first one, is 7 [#permalink]

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New post 02 Aug 2017, 07:16
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Bunuel wrote:
In a sequence of 40 numbers, each term, except for the first one, is 7 less than the previous term. If the greatest term in the sequence is 281, what is the smallest term in the sequence?

A. 8
B. 7
C. 1
D. 0
E. −6


Common difference, d = -7
Since, it is a decreasing arithmetic progression, the greatest term is the first term and last term is smallest .
a1 = 281
Last term = a1 + 39*-7
= 281 -273 = 8

Answer A
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Re: In a sequence of 40 numbers, each term, except for the first one, is 7 [#permalink]

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New post 03 Aug 2017, 11:54
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Since each term in the sequence is 7 less than the previous term, thus the first term x has to be the greatest term and the last term i.e. 40th term would be smallest.
The 2nd term would be x-7 i.e. x-(1)7.
The 3rd term would be x-14 i.e. x-(2)7.
Thus the smallest term would be x-273 i.e. x-(39)7.
The greatest term in the sequence x=281.
So the smallest term would be x-273=281-273=8.
Ans A

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Re: In a sequence of 40 numbers, each term, except for the first one, is 7 [#permalink]

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New post 09 Aug 2017, 12:51
Bunuel wrote:
In a sequence of 40 numbers, each term, except for the first one, is 7 less than the previous term. If the greatest term in the sequence is 281, what is the smallest term in the sequence?

A. 8
B. 7
C. 1
D. 0
E. −6


Note that this is a decreasing sequence. Since each term except for the first term is 7 less than the previous term, and the greatest term is 281, 281 must be the first term, and the succeeding terms are 274, 267, 260, and so on.

We can see that this sequence is an arithmetic sequence in which we can use the following formula to find the smallest term, i.e., the last term, of the sequence: a_n = a_1 + (n - 1)d, where a_n is the nth term, a_1 is the first term, n is the number of terms, and d is the common difference.

Here, a_1 = 281, n = 40, and d = -7, so the smallest term is a_40 = 281 + (40 - 1)(-7) = 281 - 273 = 8.

Answer: A
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Re: In a sequence of 40 numbers, each term, except for the first one, is 7   [#permalink] 09 Aug 2017, 12:51
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In a sequence of 40 numbers, each term, except for the first one, is 7

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