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If a sequence of consecutive integers of increasing value has a sum of

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Bunuel
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If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   12 Dec 2018, 01:29
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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
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E
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Archit3110
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Re: If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   12 Dec 2018, 01:38
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Bunuel wrote:
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



sn= n/2 ( 2a+ (n-1)d)

63= n/2 ( 12+n-1)

solve for n we get n = 7

IMO E
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Chethan92
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Re: If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   12 Dec 2018, 01:39
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\(S_n = \frac{n}{2}[2a+(n-1)d]\)
\(a = 6\)
\(d = 1\)
\(S_n = 63\)
\(63 = \frac{n}{2}[12+n-1]\)
\(126 = n^2+11n\)
\(n^2+11n-126 = 0\)
\(n^2+18-7n-126 = 0\)
n = 7 or -18
n cannot be negative

E is the answer.
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KHow
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Re: If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   07 Jan 2019, 06:40
Hi Bunuel,

Could you please provide your feedback on how to solve this problem? I am stuck on C; however, the correct answer is E.

Thank you!

Bunuel wrote:
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
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gracie
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If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   07 Jan 2019, 12:36
Bunuel wrote:
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


because 63 sum=number of terms*mean,
try choices 9 and 7 as factors of 63
7 terms with mean of 9 works
E
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Zed3092
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Re: If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   16 Jan 2019, 02:29
Afc0892 wrote:
\(S_n = \frac{n}{2}[2a+(n-1)d]\)
\(a = 6\)
\(d = 1\)
\(S_n = 63\)
\(63 = \frac{n}{2}[12+n-1]\)
\(126 = n^2+11n\)
\(n^2+11n-126 = 0\)
\(n^2+18-7n-126 = 0\)
n = 7 or -18
n cannot be negative

E is the answer.


Could you please explain why have you taken D=1?
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Chethan92
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Re: If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   16 Jan 2019, 02:45
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nehalkapadi123 wrote:
Afc0892 wrote:
\(S_n = \frac{n}{2}[2a+(n-1)d]\)
\(a = 6\)
\(d = 1\)
\(S_n = 63\)
\(63 = \frac{n}{2}[12+n-1]\)
\(126 = n^2+11n\)
\(n^2+11n-126 = 0\)
\(n^2+18-7n-126 = 0\)
n = 7 or -18
n cannot be negative

E is the answer.


Could you please explain why have you taken D=1?


Consecutive integers have a difference of 1. Hence D = 1. :)

Posted from my mobile device
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Zed3092
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Re: If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   16 Jan 2019, 02:48
Afc0892 wrote:
nehalkapadi123 wrote:
Afc0892 wrote:
\(S_n = \frac{n}{2}[2a+(n-1)d]\)
\(a = 6\)
\(d = 1\)
\(S_n = 63\)
\(63 = \frac{n}{2}[12+n-1]\)
\(126 = n^2+11n\)
\(n^2+11n-126 = 0\)
\(n^2+18-7n-126 = 0\)
n = 7 or -18
n cannot be negative

E is the answer.


Could you please explain why have you taken D=1?


Consecutive integers have a difference of 1. Hence D = 1. :)

Posted from my mobile device


Damn, I completely missed the term consecutive. Thank you
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Re: If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   17 Jan 2019, 10:44
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Archit3110 Afc0892

Hey guys I know that the general formula is (Afist+A final)n/2 how did u come up with this one sn= n/2 ( 2a+ (n-1)d) ?
I would really appreciate some help, thank you in advance
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Re: If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   17 Jan 2019, 10:49
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UNSTOPPABLE12 wrote:
Archit3110 Afc0892

Hey guys I know that the general formula is (Afist+A final)n/2 how did u come up with this one sn= n/2 ( 2a+ (n-1)d) ?
I would really appreciate some help, thank you in advance


UNSTOPPABLE12
its a general formula to determine sum of no in a sequence .
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Re: If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   17 Jan 2019, 10:59
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KHow

An another easy way to solve this without using formulas would just be to calculate that is the way I did it.

6 is the starting point, and you know that the numbers are consecutive, fortunately the sum is really low (63) thus calculating is feasible within a small time frame.

6+7+8+9+10+11+12=63, thus the answer is 7
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If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   07 Apr 2020, 20:06
Bunuel wrote:
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


let n=number of terms
sum of first and last terms=6+[6+(n-1)]→11+n
sum of sequence 63=n(11+n)/2→
n=7 terms
E
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Re: If a sequence of consecutive integers of increasing value has a sum of [#permalink] New post   08 Apr 2021, 10:25
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Re: If a sequence of consecutive integers of increasing value has a sum of [#permalink]
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