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# When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu

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Math Revolution GMAT Instructor
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When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu  [#permalink]

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Updated on: 09 Feb 2018, 01:34
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68% (02:09) correct 32% (02:59) wrong based on 59 sessions

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When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive multiples of 6 to 714, inclusively?

A. 3*119*120
B. 6*119*120
C. 3*118*119
D. 6*120*121
E. 3*120*121

*A solution will be posted in two days.

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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Originally posted by MathRevolution on 11 Jan 2016, 16:54. Last edited by MathRevolution on 09 Feb 2018, 01:34, edited 1 time in total. CEO Joined: 12 Sep 2015 Posts: 3728 Location: Canada Re: When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu [#permalink] ### Show Tags 11 Jan 2016, 17:02 MathRevolution wrote: When 1+2+..........+n = n(n+1)/2, what is the sum of all the positive multiples of 6 to 714, inclusively? A. 3*119*120 B. 6*119*120 C. 3*118*119 D. 6*120*121 E. 3*120*121 We want the sum of: 6 + 12 + 18 + 24 + ... + 708 + 714 Factor out a 6 to get: 6(1 + 2 + 3 + 4 + ... + 118 + 119) Apply given formula to get: 6[(119)(119 + 1)/2] Simplify: 6[(119)(120)/2] Simplify more to get: (3)(119)(120) Answer: A Cheers, Brent _________________ Test confidently with gmatprepnow.com Intern Joined: 28 Sep 2011 Posts: 8 Re: When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu [#permalink] ### Show Tags 13 Jan 2016, 08:25 1 MathRevolution wrote: When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive multiples of 6 to 714, inclusively? A. 3*119*120 B. 6*119*120 C. 3*118*119 D. 6*120*121 E. 3*120*121 *A solution will be posted in two days. f(n) = 1 + 2 + ... + n = n(n+1)/2 714 = 6 x 119 and multiples of 6 to 714 include: 6x1, 6x2, 6x3, ....6x119 Sum of all multiples = 6 x f(119) = 6 x 119 x 120/2 = 3x119x120 Answer is A Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7372 GMAT 1: 760 Q51 V42 GPA: 3.82 When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu [#permalink] ### Show Tags Updated on: 09 Feb 2018, 14:13 When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive multiples of 6 to 714, inclusively? A. 3*119*120 B. 6*119*120 C. 3*118*119 D. 6*120*121 E. 3*120*121 -> 6+12+18+….+708+714=6(1+2+3+…+118+119) =6[119(119+1)]/2=3*119*120. Therefore, the answer is A. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Originally posted by MathRevolution on 13 Jan 2016, 19:06.
Last edited by MathRevolution on 09 Feb 2018, 14:13, edited 2 times in total.
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Re: When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu  [#permalink]

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08 Sep 2017, 10:17
First lets calculate no of terms there are from 6 to 714 that are multiples of 6
So
general formula for counting in steps of x is ( LT-FT)/x
If both ends are to be included add+1 or if both ends are excluded -1. If only one end taken then formula is as is.
Counting in steps of 6 We can write no terms as{ (714-6)/6}+1 = 119
Now since the seqn. will be in AP sum is 119 *{(6+714) /2}
= 119 *360= 3*119*120
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Re: When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu  [#permalink]

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09 Sep 2017, 16:15
1
MathRevolution wrote:
When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive multiples of 6 to 714, inclusively?

A. 3*119*120 B. 6*119*120 C. 3*118*119 D. 6*120*121 E. 3*120*121

*A solution will be posted in two days.

Half the challenge might be wading through answers as formatted.

For sums of multiples of sequences, I do not use factoring or n(n-1)/2.

I used the general formula for summing [multiples of] an arithmetic sequence, then found the value that matched my answer.

Formula for sum of multiples of arithmetic sequences:

(Average)(# of terms), where

Average = $$\frac{FirstTerm + Last Term}{2}$$ and

# of Terms: $$\frac{Last - First}{Increment}$$ + 1

Average: $$\frac{6 + 714}{2}$$ = 360

# of Terms: $$\frac{708}{6}$$ = 118 + 1 = 119

Sum of multiples of 6, from 6 to 714 inclusive, is (360)(119)

Check answers (and toss out B and D immediately, they're huge, and C, which does not end in a zero).

Answer A is a match.
3*119*120 = (119)(360)

ANSWER A
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Re: When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu   [#permalink] 09 Sep 2017, 16:15
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# When 1+2+..........+n=n(n+1)/2, what is the sum of all the positive mu

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