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M03-29

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New post 16 Sep 2014, 00:21
1
7
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A
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D
E

Difficulty:

  15% (low)

Question Stats:

84% (02:11) correct 16% (02:49) wrong based on 103 sessions

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The sequence \(a_1\), \(a_2\), ..., \(a_n\), ... is such that \(a_n=a_1+(n-1)d\) for all integers \(n \gt 1\) and for some constant \(d\). If the sum of the second and the fifth terms of the sequence is 8 and the sum of the third and the seventh terms is 14, what is the first term of the sequence?

A. 3
B. 2
C. 1
D. -1
E. -3

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New post 16 Sep 2014, 00:21
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Official Solution:

The sequence \(a_1\), \(a_2\), ..., \(a_n\), ... is such that \(a_n=a_1+(n-1)d\) for all integers \(n \gt 1\) and for some constant \(d\). If the sum of the second and the fifth terms of the sequence is 8 and the sum of the third and the seventh terms is 14, what is the first term of the sequence?

A. 3
B. 2
C. 1
D. -1
E. -3


Given:

\(a_2+a_5=(a_1+d)+(a_1+4d)=2a_1+5d=8\);

\(a_3+a_7=(a_1+2d)+(a_1+6d)=2a_1+8d=14\);

Subtract 1 from 2: \(3d=6\), hence \(d=2\). Now, since \(2a_1+5d=8\) then \(2a_1+5*2=8\) giving \(a_1=-1\).


Answer: D
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New post 09 Jun 2015, 06:00
I solved it in a different way, mostly applying mental & logical sense.

1st case = 2nd + 5th = 8
2nd case = 3rd + 7th = 14
--------------------------
Differnce= 1d + 2d = 6

So, d = 2 or sequence difference is 2.

Now, for second case, actual difference is 7th-3rd = 4d, hence 4d*2 = 8.

Now we can think what 2 numbers, whose difference is 8, sum up to 14?

If we just plug&play, the numbers will come easily= 3 & 11

If the 3rd number is 3, then 1st number is = 3-2-2 = -1 [D is correct] :)
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Re: M03-29  [#permalink]

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New post 09 Jun 2015, 07:01
Solved it correctly applying Bunuel's approach. My problem is - I needed approx. 2 mins for this. Should I have been faster or is this normal for a problem involving 3-4 operations?

Thanks!
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New post 09 Jun 2015, 07:25
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New post 31 Oct 2015, 20:48
a2=a1+d
a5=a1+4d
2a1 + 5d = 8 - multiply by 8

a3 = a1+2d
a7=a1+6d
2a1+8d=14 - multiply by 5


16a1+40d = 64_
10a1+40d = 70
6a1 = -6
a1 = -1
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New post 13 Feb 2016, 09:17
Sorry, one question.
it's clear that d=2 and a1=-1. Tho the exercise starts saying that the equation is valid for all integers n > 1. therefore, the first term of the sequence should be 1 and not -1. a1= a1 + (n-1)d= -1. a2=a2+(n-1)d=1. If the correct answer is -1 than the equation should be valid for all n>=1
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New post 17 Feb 2016, 21:24
Talderigi wrote:
Sorry, one question.
it's clear that d=2 and a1=-1. Tho the exercise starts saying that the equation is valid for all integers n > 1. therefore, the first term of the sequence should be 1 and not -1. a1= a1 + (n-1)d= -1. a2=a2+(n-1)d=1. If the correct answer is -1 than the equation should be valid for all n>=1



only "n" has to be greater than 1

However the "function", a(n) can equal anything

In this case a(1) = -1

You would not be allowed to have a(-1) = whatever
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New post 22 Sep 2018, 19:58
I have a question. Since it states that n>1, then shouldn't the first term be a2 = a1 + d ? Since there is no a1 (which means that n=1 which violates n>1). Therefore, the second term stated in the question should be a3 = a1 + 2d?
Please correct my misunderstanding.
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New post 23 Sep 2018, 03:53
new30 wrote:
I have a question. Since it states that n>1, then shouldn't the first term be a2 = a1 + d ? Since there is no a1 (which means that n=1 which violates n>1). Therefore, the second term stated in the question should be a3 = a1 + 2d?
Please correct my misunderstanding.


The sequence \(a_1\), \(a_2\), ..., \(a_n\), ... is such that \(a_n=a_1+(n-1)d\) for all integers \(n \gt 1\) and for some constant \(d\).

The above means that in the given sequence, formula for nth term (\(a_n=a_1+(n-1)d\)) can be applied for all terms starting from \(a_2\). So, \(a_2=a_1+(2-1)d\),
\(a_3=a_1+(3-1)d\), ...
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Re: M03-29  [#permalink]

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New post 13 Jun 2019, 07:13
Bunuel wrote:
The sequence \(a_1\), \(a_2\), ..., \(a_n\), ... is such that \(a_n=a_1+(n-1)d\) for all integers \(n \gt 1\) and for some constant \(d\). If the sum of the second and the fifth terms of the sequence is 8 and the sum of the third and the seventh terms is 14, what is the first term of the sequence?

A. 3
B. 2
C. 1
D. -1
E. -3



Hi @Bunuel--- another approch
for the second equation the algebraic form becomes

a+2d + a+6d = 14
2a+8d=14

dividing by 2
a + 4d = 7 (which becomes the 5th term)

so as per first equation

IInd term + Vth term = 8
IInd term = 1

2nd term = 1
3rd term = 3
4th term = 5
5th term = 7

d = 2

so first term is -1
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Re: M03-29   [#permalink] 13 Jun 2019, 07:13
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