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Sum of all terms in the set {S13, S14, ..., S28}?

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Sum of all terms in the set {S13, S14, ..., S28}?

by gmattesttaker2 » Wed May 14, 2014 11:15 pm
Hello,

Can you please tell me what is the best approach to solve this problem:

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

OA: 1,968


My approach:

S13 = 78
S28 = 168

Number of terms, n = ( 28 - 13 ) + 1 = 16

Average = Sum/n

=> (78 + 168)/2 = Sum/16
=> 246/2 = Sum/16
=> 123 = Sum/16
=> Sum = 123 x 16

Since only D has a unit's digit ending in 8, answer is D

I was just wondering if there is a quicker approach to solving this problem? Thanks a lot.

Regards,
Sri
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by [email protected] » Wed May 14, 2014 11:27 pm
Hi Sri,

You've used a standard, algebraic approach to this question, which is fine. Another way to answer this question is to use "bunching"....

Since we're dealing with the 13th through 28th terms, we're dealing with 16 terms....

The sum of the 1st and 16th term = 78 + 168 = 246
The sum of the 2nd and 15th term = 84 + 162 = 246
etc.

So, we have 8 "sets" of 2 terms that all sum to 246

8(246) = 1968

The "shortcut" that you mentioned about the unit's digit still applies here.

Final Answer: D

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by sanju09 » Thu May 15, 2014 1:08 am
gmattesttaker2 wrote:Hello,

Can you please tell me what is the best approach to solve this problem:

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

OA: 1,968


My approach:

S13 = 78
S28 = 168

Number of terms, n = ( 28 - 13 ) + 1 = 16

Average = Sum/n

=> (78 + 168)/2 = Sum/16
=> 246/2 = Sum/16
=> 123 = Sum/16
=> Sum = 123 x 16

Since only D has a unit's digit ending in 8, answer is D

I was just wondering if there is a quicker approach to solving this problem? Thanks a lot.

Regards,
Sri
Shortcut maybe the fact that the average (arithmetic mean) of evenly spaced numbers is equal to the average of first and last numbers in the list, which is a set of 16 consecutive multiples of 6 with the first term 78 and the last term 168, their average 39 + 84, sum of all is number times average, which is 16(123), and here you are perfect why D.

In fact, your approach is the shortcut!
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by Matt@VeritasPrep » Fri May 16, 2014 12:02 pm
One time-saving device here: since you have an EVEN number of evenly spaced terms in the set, you don't need to average them, you can just treat them as pairs.

For instance, take the set {2, 4, 6, 8, 10, 12}

Notice that 2 + 12 = 4 + 10 = 6 + 8, so the sum is just 14 * 3.

Similarly, in your set, you'll have 8 pairs, each with a sum of (78+168), so your sum is 8*(78+168), or 8*(250-4) = 2000 - 32 = 1968.
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by GMATGuruNY » Sat May 17, 2014 2:19 am
gmattesttaker2 wrote:Hello,

Can you please tell me what is the best approach to solve this problem:

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
For any EVENLY SPACED SET:
Sum = (number)(median).

From S�₃ to S₂₈:
Number of terms = 16.
The median is HALFWAY between S₂₀ and S₂�.
S₂� = S� + 20(6) = 6 + 20(6) = 126.
S₂₀ = S₂� - 6 = 126-6 = 120.
Median = halfway between 120 and 126 = 123.
Sum = (number)(median) = 16*123 = product with a units digit of 8.

The correct answer is D.
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