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Series T is a sequence of numbers where each term after the

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Series T is a sequence of numbers where each term after the [#permalink] New post 30 Dec 2010, 10:53
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Series T is a sequence of numbers where each term after the first term is x greater than the term that precedes it. If the sum of the first and last terms of series T is 14, then what is the sum of the first three terms of series T and the last three terms of series T?

A. -7
B. 7
C. 14
D. 42
E. 84
[Reveal] Spoiler: OA
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Re: Kaplan Sequence Teaser [#permalink] New post 30 Dec 2010, 11:21
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ezinis wrote:
Series T is a sequence of numbers where each term after the first term is x greater than the term that precedes it. If the sum of the first and last terms of series T is 14, then what is the sum of the first three terms of series T and the last three terms of series T?
A) -7
B) 7
C) 14
D) 42
E) 84


T is an evenly spaced set. For evenly spaced set a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=..., so if given that a_1+a_n=14 then (a_1+a_n)+(a_2+a_{n-1})+(a_3+a_{n-2})=3*14=42.

Answer: D.
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Re: Kaplan Sequence Teaser [#permalink] New post 30 Dec 2010, 12:07
Bravo, I found my mistake thanks to you. I was obsessing about "x times greater ...". But I solved it. So lets pretend the stem says ... x times greater ... Can someone give a solution to approach this?
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Re: Kaplan Sequence Teaser [#permalink] New post 30 Dec 2010, 13:33
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ezinis wrote:
Bravo, I found my mistake thanks to you. I was obsessing about "x times greater ...". But I solved it. So lets pretend the stem says ... x times greater ... Can someone give a solution to approach this?


You won't get the answer in this case, as you can not calculate: a+ax+ax^2+ax^{n-3}+ax^{n-2}+ax^{n-1} just knowing that a+ax^{n-1}=14
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Re: Kaplan Sequence Teaser [#permalink] New post 08 Jan 2011, 22:20
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my way of solving :P
a+l = 14
now the series would be a,a+d,a+2d...........l-2d,l-d,l
now the sum of the first three terms and last three terms would be a+a+d+a+2d+l-2d+l-d+l
which comes out to be 3 (a+l) = 3 (14) = 42
I hope this is the easiest method of solving the question... :lol: :lol:
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Re: Kaplan Sequence Teaser [#permalink] New post 28 Sep 2012, 04:15
This is the way i solved it ..


The question gives us a general formula that confirms that we are dealing with an A.P. , therefore we know that the nth term = a (n-1)d .. In this case d = x ...

First term of the AP = A ; Last term = L ..

we know that A + L = 14 ...........[I]

Now we are to find the sum of 6 terms , ie the first three and the last three ... Lets assume that this series has 6 numbers .. therefore Sum of 6 numbers can be written as

S6 = 6/2 [ A + L ]

= 3 x 14 ( From I)

= 42 (D)


As an alternative to this method , we could PLUG IN values for X and solve for the sum of 6 terms of the AP as follows ------------->

Let us assume x =1 , therefore we would need to come up with a sequence whose first and last term adds to 14 , and which as at least 6 terms ...

let us use this sequence ...

3 , 4 , 5, 6, 7, 8, 9, 10, 11

The first and the last terms add to 14 ...

Each term is x greater than the one it succeeds ( in this case i have assumed x to be 01)

The first three and the last three terms are : 3 + 4 + 5 + 9 + 10 + 11 = 42 (D)


To further test the plug in method (not required to confirm the answer but just to test it out) ...

Now let us assume that x = 2

1 3 5 7 9 11 13

First three and last three terms add up to = 42 (D) ....
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Re: Series T is a sequence of numbers where each term after the [#permalink] New post 08 Dec 2013, 04:41
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Re: Series T is a sequence of numbers where each term after the   [#permalink] 08 Dec 2013, 04:41
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