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# If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn =

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CEO
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If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = [#permalink]

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15 Feb 2008, 14:34
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If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = Sn-1 + 6,..., what is the sum of all terms in the set {S13, S14, ..., S28}?
1,800
1,845
1,890
1,968
2,016
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You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

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Director
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15 Feb 2008, 14:57
bmwhype2 wrote:
If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = Sn-1 + 6,..., what is the sum of all terms in the set {S13, S14, ..., S28}?
1,800
1,845
1,890
1,968
2,016

sum = ((s13+s28)/2)*(28-13+1) = (s13+s28)*8 = (6 + 12*6+6+27*6)*8 = 6*8*(2+27+12)=6*8*41 = 1968 -> D

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SVP
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15 Feb 2008, 19:32
i started going this way, but got stuck .. i must be missing something basic.

Sn=6*n ... so S1=6, S2=12 and so on.

Sum of terms from S13 to S28 can be written as : 6*(13+14+...+28). Now, where can I go from here ?

edit: oops, just got it. I think theres a formula you use for sum of consecutive terms...

(last term + first term)/2 * (number of terms in sequence)

What cases does this formula apply in ?

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CEO
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16 Feb 2008, 00:39
2
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bmwhype2 wrote:
If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = Sn-1 + 6,..., what is the sum of all terms in the set {S13, S14, ..., S28}?
1,800
1,845
1,890
1,968
2,016

Basically its just (n-1)*6+6 so S13 is (12)6+6 --> 78

S28 is (27)6+6 --> 168

Now there are 16 terms from 13 to 28 and the average is (168+78)/2 -->

Now just 16*123 --> 1968

D

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Director
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16 Feb 2008, 10:40
Basically need to realize that each increment will be times 6.

For example

The set starting at 1 is:
x, 2x, 3x, 4x, 5x... where x = 6.

So we need to sum 13x --> 28x. We just use the summation formula:

(16/2) * 41 = 328x = 328(6) = 1968 D.

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Senior Manager
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17 Feb 2008, 12:33
+1 to GMATBLACKBELT....really simple less than 2min approach....
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CEO
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19 Feb 2008, 08:36
GMATBLACKBELT wrote:
bmwhype2 wrote:
If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = Sn-1 + 6,..., what is the sum of all terms in the set {S13, S14, ..., S28}?
1,800
1,845
1,890
1,968
2,016

Basically its just (n-1)*6+6 so S13 is (12)6+6 --> 78

S28 is (27)6+6 --> 168

Now there are 16 terms from 13 to 28 and the average is (168+78)/2 -->

Now just 16*123 --> 1968

D

thanks. this is why i post questions that i know.

i did the entire problem via AP and then sum of AP, it took 3 minutes to solve.
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You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

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Director
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20 Feb 2008, 01:27
1
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I did it in 60 seconds.

Here's my way

Rewrite Sn as S(n) = 6*n.

So we want 6(13) + 6(14) + ... + 6(28)
= 6 (13 + 14 + ... + 28 )
= 6 ( sum of first 28 numbers - sum of first 12 numbers )
= 6 ( 28*29/2 - 12*13 / 2 )
= 3 ( 812 - 156 )
= 3 ( 656 )
= 1968

D)

Good question

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Re: MGMAT - Sequence   [#permalink] 20 Feb 2008, 01:27
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# If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn =

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