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

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by Jake@ThePrincetonReview » Thu Mar 29, 2018 11:38 am
An easy way to find the sum of consecutive numbers is to recognize that, in a list of consecutive numbers, the average is equal to the median. So all you have to do is find the median and multiply that by the number of numbers to get the sum.

So for this problem, first figure out the number of numbers. From S13 to S28, there are 16 numbers. (28-13) + 1.
Next, find the median. With an even number of numbers, the median will the average of the two numbers in the middle, S20 and S21.
Noticing that S1 = 6 x 1, S2 = 6 x 2 etc..., S20 will be 6 x 20 = 120 and S21 will be 6 x 21 = 126, so their average is 123.
Last step, multiply 123 (the median) by the number of numbers (16) to get 1968, D.

Bonus time earned if you noticed that the units digit of 123 x 16 will be an 8, and only one answer choice ends in an 8.

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by Scott@TargetTestPrep » Fri Mar 30, 2018 11:11 am
BTGmoderatorRO 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}?

A. 1,800
B. 1,845
C. 1,890
D. 1,968
E. 2,016

We see that the sequence is:

6, 12, 18, 24, ...

This is an arithmetic sequence with first term S1 = 6 and common difference d = 6.

Recall that S_n = S1 + d(n - 1), so S13 = 6 + 6(13 - 1) = 78 and S28 = 6 + 6(28 - 1) = 168. To find the sum of a list of consecutive terms of an arithmetic sequence, we can use the formula:

(number of terms) x (first term + last term)/2

Here, the number of terms = 28 - 13 + 1 = 16, the first term = S13 = 78, and the last term = S28 = 168; thus; the sum of the terms form S13 to S28, inclusive, is:

16 x (78 + 168)/2

16 x 123

1,968

Answer: D

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by deloitte247 » Sat Mar 31, 2018 12:50 pm
S1=6
S2=12
$$So,\ S13=13\cdot6=78$$
$$S28=28\cdot6=168$$
$$The\ sum\ of\ all\ terms\ in\ the\ set=number\ of\ term\cdot\frac{\left(1st\ term+lost\ term\right)}{2}$$
$$=16\cdot6\cdot\frac{\left(13+28\right)}{2}$$
$$=96\cdot\frac{41}{2}=\frac{3936}{2}=1968\ \left(option\ D\right)$$
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