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# M03-29

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Math Expert
Joined: 02 Sep 2009
Posts: 52431

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15 Sep 2014, 23:21
1
7
00:00

Difficulty:

15% (low)

Question Stats:

84% (01:33) correct 16% (02:33) wrong based on 162 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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15 Sep 2014, 23:21
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1
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 $$a_1=-1$$.

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Joined: 22 Dec 2007
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09 Jun 2015, 05:00
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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09 Jun 2015, 06:01
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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Thank you very much for reading this post till the end! Kudos?

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09 Jun 2015, 06:25
1
bgpower wrote:
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!

I'd say any time less than 2 minutes is fine.
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31 Oct 2015, 19:48
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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13 Feb 2016, 08:17
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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17 Feb 2016, 20:24
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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22 Sep 2018, 18:58
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?
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23 Sep 2018, 02:53
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?

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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06 Oct 2018, 19:05
How is this a 500 level question? I think this is 700 level.
Re: M03-29 &nbs [#permalink] 06 Oct 2018, 19:05
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# M03-29

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