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# If a sequence of consecutive integers of increasing value

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Director
Joined: 12 Jun 2006
Posts: 536
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Kudos [?]: 28 [0], given: 1

If a sequence of consecutive integers of increasing value [#permalink]  23 Oct 2006, 01:23
If a sequence of consecutive integers of increasing value has a sum of 63 and a first term of 6, how many integers are in the sequence?

(A) 11
(B) 10
(C) 9
(D) 8
(E) 7
Manager
Joined: 07 Jun 2006
Posts: 113
Followers: 2

Kudos [?]: 7 [0], given: 0

[#permalink]  23 Oct 2006, 01:55
E. 7

63 = (n/2) (2*6+(n-1)*1)

Where n is the number of terms.....

Solving for n, we get n = -18 OR n = 7

Verifying , 6+7+8+9+10+11+12 = 63
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1

Kudos [?]: 17 [0], given: 0

[#permalink]  23 Oct 2006, 02:25
Girikorat, did you set-up that equation or is that a formula you keep in mind? If it's a formula, can you state in general terms? Thanks
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Director
Joined: 12 Jun 2006
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Kudos [?]: 28 [0], given: 1

[#permalink]  23 Oct 2006, 16:40
Yes, please explain.

Also, the problem says "consecutive integers". Will this always mean 6,7,8,9 ... or might they mean consecutive even/odd numbers?
Manager
Joined: 28 Aug 2006
Posts: 160
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Kudos [?]: 13 [0], given: 0

[#permalink]  23 Oct 2006, 18:02
Sum of n terms in an arithmetic series is

N/2(2a+(N-1)d)

where N-- number of terms
a--- 1st term
d----difference.

Also FYI--

N th term is given by

a+(n-1)d
Manager
Joined: 07 Jun 2006
Posts: 113
Followers: 2

Kudos [?]: 7 [0], given: 0

[#permalink]  23 Oct 2006, 20:22
Code:
Sum of n terms in an arithmetic series is

N/2(2a+(N-1)d)

where N-- number of terms
a--- 1st term
d----difference.

Also FYI--

N th term is given by

a+(n-1)d

Nothing to add
[#permalink] 23 Oct 2006, 20:22
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# If a sequence of consecutive integers of increasing value

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