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# What is the tenth term in a series of 89 consecutive

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VP
Joined: 30 Jun 2008
Posts: 1026
What is the tenth term in a series of 89 consecutive [#permalink]

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09 Oct 2008, 07:37
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What is the tenth term in a series of 89 consecutive positive integers?
1) The average of the integers is 100
2) The 79th term in the series is 134
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Intern
Joined: 09 Oct 2008
Posts: 23
Re: DS : Consecutive Integers [#permalink]

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09 Oct 2008, 08:20

The question is solvable if you know ANY one integer with its position between 1st to 89th consecutive number. Then you can find any number in any position because they are consecutive series.

1) Average in consecutive numbers means the position is exactly in the middle from 1st to 89th number. (Don't even bother trying to calculate). The point is that you know exact number on a exact position.
(SUFFICENT)

2) Exactly what I was looking from above.
(SUFFICIENT)

Last edited by cityjoy92 on 09 Oct 2008, 08:41, edited 3 times in total.
SVP
Joined: 29 Aug 2007
Posts: 2467
Re: DS : Consecutive Integers [#permalink]

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09 Oct 2008, 08:23
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amitdgr wrote:
What is the tenth term in a series of 89 consecutive positive integers?

1) The average of the integers is 100
2) The 79th term in the series is 134

D.

1: average is 100, then 100 is 45th term. so 1st term is 56, 10th term is 64 and 89th or last term is 144.
2: similar approach.
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Senior Manager
Joined: 04 Jan 2006
Posts: 276
Re: DS : Consecutive Integers [#permalink]

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09 Oct 2008, 08:46
amitdgr wrote:
What is the tenth term in a series of 89 consecutive positive integers?
1) The average of the integers is 100
2) The 79th term in the series is 134

An = A1 + d(n-1)
d = 1 ---------- because of consecutive integers
An = A1 + n - 1
An - A1 = n - 1 ---------- (1)

1) Ave(Sn) = 100
n = 89
(A1 + A89) / 2 = 100
A89 + A1 = 200
A89 - A1 = 89 - 1 = 88 ---------- From (1)

Equation can be solve. Know A10. Sufficient.

2) A79 = 134
A79 - A1 = 79 - 1 = 78 ---------- From (1)
134 - A1 = 78
If we know A1, A10 can be solved.
Sufficient.

Last edited by devilmirror on 09 Oct 2008, 09:20, edited 1 time in total.
VP
Joined: 30 Jun 2008
Posts: 1026
Re: DS : Consecutive Integers [#permalink]

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09 Oct 2008, 08:59
Thanks guys. I was comfortable with statement 2. But I missed the point that 100 has to be the middle term.

DevilMirror .. That is a nice method to solve

Thanks guys, OA is D
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Re: DS : Consecutive Integers [#permalink]

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09 Oct 2008, 09:00
My two cents...

1 because they are consecutive, we can write:
x+(x+1)+(x+2)+...+(x+88)
If the mean in known, we can calculate x+9 (the tenth term)
SUFF.

2 again, the 79th term can be written as x+78. Thus, we know x+9 (which is the tenth term)
SUFF.

IMO: D

OA?

Cheers
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VP
Joined: 30 Jun 2008
Posts: 1026
Re: DS : Consecutive Integers [#permalink]

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09 Oct 2008, 18:08
JohnLewis1980 wrote:
My two cents...

1 because they are consecutive, we can write:
x+(x+1)+(x+2)+...+(x+88)
If the mean in known, we can calculate x+9 (the tenth term)
SUFF.

2 again, the 79th term can be written as x+78. Thus, we know x+9 (which is the tenth term)
SUFF.

IMO: D

OA?

Cheers

Thanks John. That is a nice way to approach. OA is D
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Re: DS : Consecutive Integers   [#permalink] 09 Oct 2008, 18:08
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