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Infinite sequence problem

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Infinite sequence problem

by rohan.tarun » Mon Apr 24, 2017 10:54 am
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

I recently took the MGMAT CAT and came across this question. Can anyone please help me solve this in a 2 minute scenario? I did figure out the sequence can be represented by 6n however the further part in the question is what stumped me and would like to know of an approach that can be used in more problems like this one. Thanks.
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by regor60 » Mon Apr 24, 2017 11:10 am
rohan.tarun 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

I recently took the MGMAT CAT and came across this question. Can anyone please help me solve this in a 2 minute scenario? I did figure out the sequence can be represented by 6n however the further part in the question is what stumped me and would like to know of an approach that can be used in more problems like this one. Thanks.

OK, so you've figured out that each term is represented by 6x the term number. So, you can bring the 6 out like this:

6*(13+14+15...28) = the desired sum

So, now you want to find the total of the summation. A trick to remember is to notice that you can write the sum as follows:

13 14 15...20
+ 28 + 27 + 26.. +21

Notice that each adds up to 41. And there are 8 such totals. So the sum is 8*41.

Don't forget then to multiply this by 6, so 6*8*41 = 1968
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by GMATGuruNY » Mon Apr 24, 2017 11:59 am
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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by [email protected] » Mon Apr 24, 2017 2:35 pm
Hi rohan.tarun,

To start, your goal should NOT be to answer each Quant question in under 2 minutes, since certain questions will require more than 2 minutes of work to answer. Your actual goal should be to deal with each question in the most efficient way possible (even if that means 'dumping' a question and moving past it).

The numbers in this question can be 'bunched' into pairs that have the same sum. According to the information in the prompt, we can determine each of the terms...

1st = 6
2nd = 12
3rd = 18

Notice that the pattern can be defined as (number of term)(6), so...
4th = 24
5th = 30
....
27th = (27)(6) = 162
28th = (28)(6) = 168

We're asked for the sum of the 13th through 28th terms, inclusive. This means that we're adding 16 terms...

If we add the 13th term and 28th term, we get 78 + 168 = 246
If we add the 14th term and 27th term, we get 84 + 162 = 246
Etc.

Since there are 16 terms, there are 8 'pairs' that total 246 each. Thus, the sum is (8)(246). You should notice that the units digit of that product is 8... and there's only one answer that fits that pattern.

Final Answer: D

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by Matt@VeritasPrep » Thu Apr 27, 2017 11:52 pm
We could also say that the sum is like so:

13*6 + 14*6 + ... + 28*6 =>

6 * (13 + 14 + ... + 28) =>

6 * ((1 + 2 + ... + 28) - (1 + 2 + ... + 12))

From here, use the handy trick that the sum of the integers from 1 to n = n * (n + 1)/2. That gives us

6 * (28 * 29/2 - 12 * 13/2) =>

6 * (14 * 29 - 6 * 13) =>

6 * 328 =>

1968
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by Matt@VeritasPrep » Thu Apr 27, 2017 11:54 pm
Another idea:

13*6 + 14*6 + ... + 28*6 =>

6 * (13 + 14 + ... + 28) =>

6 * (1 + 12 + 2 + 12 + ... + 16 + 12) =>

6 * (16*12 + 1 + 2 + ... + 16) =>

6 * (16*12 + 16*17/2) =>

6 * (192 + 136) =>

1968
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by Matt@VeritasPrep » Thu Apr 27, 2017 11:59 pm
We could also get really dirty and cheat with units digits.

13 + 14 + ... + 28 will have the same units digit as 3 + 4 + ... + 8. If we add the digits from 0 to 9, our sum is 45, so 10 + 11 + ... + 19 + 20 + 21 + ... + 29 would have units digits that sum to 45 + 45, or 90, giving us a units digit of 0.

We don't want to add 11, 12, or 29, however, so we take our 90 and subtract those other three units digits 1 + 2 + 9 from it. That leaves us with 90 - (1 + 2 + 9), or 78, or a units digit of 8.

So our sum 13 + 14 + ... + 28 will end in 8. That means

6 * (13 + 14 + ... 28) =>

6 * something that ends in 8 =>

something that ends in 8

and only D works.

This is the fastest novel solution I could think of, but we'd be in huge trouble if more than one answer ended in 8!
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