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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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11 Jan 2016, 16:54
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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.
[Reveal] Spoiler: OA

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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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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)

Cheers,
Brent
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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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13 Jan 2016, 08:25
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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

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Math Revolution GMAT Instructor
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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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13 Jan 2016, 19:06
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+4+…+118+119)=6[119(119+1)]/2=3*119*120. Therefore, the answer is A.
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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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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
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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.

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).

3*119*120 = (119)(360)

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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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