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S1 = 6, S2 = 12, ..., Sn = Sn-1 + 6,..., what is the sum of all terms in the set {S13, S14, ..., S28}?

formula = n/2(firstterm + last term)

= s13 to s28 ---> we have 16 terms so n will be = 16 first term = s13 = since term is getting added 6 to the next term, the 13th term will be = 13*6 = 78 s28 = 28*6 = 168

so the sum = n/2(first term + last term) = = > 16/2(78+168) ====> 1968

This question comes from Manhattan GMAT. I don't understand how you find the rule for this sequence. I just used the "\(S_n\)=\(S_(n-1)\)+6" to try to find the numbers in the sequence, but I was wrong. It's 6n. Once I see that that is the answer in the solution I can see it, but how can I arrive to that on my own?

If S is the infinite sequence \(S_1\) = 6, \(S_2\) = 12, ..., \(S_n\) = \(S_(n-1)\) + 6,..., what is the sum of all terms in the set {\(S_13\), \(S_14\), ..., \(S_28\)}? a) 1,800 b) 1,845 c) 1,890 d) 1,968 e) 2,016

This question comes from Manhattan GMAT. I don't understand how you find the rule for this sequence. I just used the "\(S_n\)=\(S_(n-1)\)+6" to try to find the numbers in the sequence, but I was wrong. It's 6n. Once I see that that is the answer in the solution I can see it, but how can I arrive to that on my own?

If S is the infinite sequence \(S_1\) = 6, \(S_2\) = 12, ..., \(S_n\) = \(S_(n-1)\) + 6,..., what is the sum of all terms in the set {\(S_{13}\), \(S_{14}\), ..., \(S_{28}\)}? a) 1,800 b) 1,845 c) 1,890 d) 1,968 e) 2,016

This is an arithmetic progression: 6, 12, 18, 24, 30...... (or I can say it is the multiplication table of 6) When they say S(n) = S(n - 1) + 6, they are giving you that every subsequent term is 6 more but just writing down the first few numbers you will realize that it is just the table of 6. This happens because the first term is 6 so every time you add 6, it just becomes the next number in the multiplication table of 6. How will you learn to observe such things? Just by practice!

First term - 6 Second term - 6x2 Third term - 6x3 and so on so 13th term will be 6x13 14th term will be 6x14 . . 28th term will be 6x28 I need to add 6x13 + 6x14 +....6x28 = 6(13 + 14 + ...28) 13 + 14 +..28 = Sum of first 28 terms - Sum of first 12 terms = \(\frac{28*29}{2} - \frac{12*13}{2} = 1968\)
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The formula is simply the formula of the sum of an AP. If a is the first term, d is the common difference, and n is the number of terms, then

\(S = \frac{n}{2}(2a + (n-1)d)\) or \(S = \frac{n}{2}(a + b)\) b is the last term of the progression which is written as a + (n-1)d. The logic behind it is that take the average of the AP which is (a + b)/2 and multiply it by n, the number of terms as if the average in added n times rather than individual numbers. It makes complete sense. Look at the example:

AP with 3 terms: 4 7 10 7 is the average. 4 is 3 less than 7 and 10 is 3 more. Rather than adding 4 and 10 to 7, I can add 7 two more times and still get the same answer.

Since GMAT does not focus on formulas, generally you can solve the question in other ways too (like I have done in my solution). Of course some basic formulas you should be good with and Sum of n consecutive terms starting from 1 = n(n + 1)/2 is one of them.
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There is a much easier way to deal with this problem. a)13th member is equal to 13*6 = smth8 (ok, 78, but 7 does not matter) b) how many members are there between 28th and 14th members (i.e how many members will we add to 13th member?) = (28-14)+1 = 15 member. c) 15*6 = smth0 That is important: the sum of members 14th-28th will has the unit digit ZERO! d) now... smth8+smth0 = smth8 or answer D in that case

I do not know where I am making mistake. Can some one please help me....

The sum of all terms in the set {S13, S14, ..., S28} means the sum of all the terms from S13 to S28, inclusive. So, it equals to the sum of first 28 terms minus the sum of first 12 terms;

Hence it should be: the sum of first 28 terms minus the sum of first 12 terms = 2436-468=1968.

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

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

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

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01 Jun 2016, 11:44

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