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# Each term of a certain sequence is 3 less than the previous term. The

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Each term of a certain sequence is 3 less than the previous term. The [#permalink]

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02 Dec 2014, 08:38
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45% (medium)

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74% (17:18) correct 26% (03:13) wrong based on 113 sessions

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Each term of a certain sequence is 3 less than the previous term. The first term of this sequence is 19. If the sum of the first n terms of the sequence is n, what is the value of positive integer n?

A 1
B 13
C 15
D 19
E 47
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Joined: 02 Sep 2009
Posts: 46035
Re: Each term of a certain sequence is 3 less than the previous term. The [#permalink]

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02 Dec 2014, 09:26
1
2
anceer wrote:
Each term of a certain sequence is 3 less than the previous term. The first term of this sequence is 19. If the sum of the first n terms of the sequence is n, what is the value of positive integer n?

A 1
B 13
C 15
D 19
E 47

We have the following evenly spaced set (aka arithmetic progression): 19, 16, 13, 10, 7, 4, 1, -2, -5, -8, -11, -14, -17, ... The first term is 19 and the common difference, d, of successive members is -3,

The sum of the elements in any evenly spaced set is given by: $$Sum=\frac{a_1+a_n}{2}*n$$, the mean multiplied by the number of terms.

Thus, we are given that $$\frac{a_1+a_n}{2}*n=n$$ --> $$a_1+a_n=2$$ --> since also given that $$a_1=19$$, then $$a_n=-17$$.

The $$n_{th}$$ term of the sequence is given by $$a_ n=a_1+d(n-1)$$ --> $$-17=19-3(n-1)$$ --> $$n=13$$.

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Re: Each term of a certain sequence is 3 less than the previous term. The [#permalink]

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02 Dec 2014, 21:43
anceer wrote:
Each term of a certain sequence is 3 less than the previous term. The first term of this sequence is 19. If the sum of the first n terms of the sequence is n, what is the value of positive integer n?

A 1
B 13
C 15
D 19
E 47

Another method is to ballpark it.
We know that it is a decreasing sequence i.e. the terms keep decreasing till 0 and then negative terms start.

What does this imply? "If the sum of the first n terms of the sequence is n"
Since number of terms will definitely be positive, we are looking for a positive sum.

19 + 16 + 13 + 10 + 7 + 4 + 1 -2 -5 -7 -10 .... and so on
Note that the first 7 terms are positive and all others negative. Every negative term has greater absolute value than the corresponding positive terms i.e. -2 absolute value is greater than 1 absolute value, -5 absolute value is greater than 4 absolute value, and so on...
Since we have 7 positive terms, we must have less than 7 negative terms to get the sum as positive. If we have 6 negative terms, we will have a total of 13 terms. Of the given options, only 13 is possible and hence it must be the answer.

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Re: Each term of a certain sequence is 3 less than the previous term. The [#permalink]

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02 Dec 2014, 22:08
19 .... 16.... 13 .......... 10 ......... 7.......... 4.......... 1 .......... -2 .......... -5 .......... -8 ........... -11 .......... -14............. -17

6 terms to the left of 1

6 terms to the right of 1

Total = 6+6+1 = 13

Corresponding terms addition = 4-2 = 7-5 = 10-8 = 2 .......... so on

Total = 1 + 2*6 = 13

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Re: Each term of a certain sequence is 3 less than the previous term. The [#permalink]

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09 Apr 2016, 19:04
anceer wrote:
Each term of a certain sequence is 3 less than the previous term. The first term of this sequence is 19. If the sum of the first n terms of the sequence is n, what is the value of positive integer n?

A 1
B 13
C 15
D 19
E 47

this one can be solved logically as well..
1 definitely can't be
47 as well..as it would be way to down in the negative side..and thus n can't be negative.

now..
19, 16, 13, 10, 7, 4, 1, -2, -5, -8, -11, -14, -17, -20, -23... list all these numbers..to go further is pointless as the sum would be negative...
19-17=2
16-14=2
13-11=2
10-8=2
7-5=2
4-2=2
1
13 numbers, sum is 13. so B it is.
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Re: Each term of a certain sequence is 3 less than the previous term. The [#permalink]

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07 Oct 2017, 01:23
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Re: Each term of a certain sequence is 3 less than the previous term. The   [#permalink] 07 Oct 2017, 01:23
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