If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = Sn-1 + 6

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

First of all, notice that we have an evenly spaced set with the first term equal to 6, and the common difference of 6: 6, 12, 18, 24, ...

Given:

\(s_1=6\);
\(s_n=s_{n-1}+6=s_1+6(n-1)\).

Question:

\(s_{13}+s_{14}+...+s_{28}=?\)

Since \(s_n=s_1+6(n-1)\) then:

\(s_{13}=6+6(13-1)=78\);
\(s_{28}=6+6(28-1)=168\).

Sum of 16 evenly spaced terms would be:

\(\frac{first \ term \ + \ last \ term}{2}*number \ of \ terms=\frac{s_{13}+s_{28}}{2}*16=\frac{78+168}{2}*16=1968\).

Answer: D.
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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

are we supposed to know this formula for GMAT?

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 = \(\frac{28*29}{2} - \frac{12*13}{2} = 1968\)

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.

Can we solve the below sum using this approach

S13-S28 = {S1-S28} - {S1-S13}

S1-S28 = n/2 {2a+(n-1)d} =28/2 { 2*6 + 27*6} = 2436
S1-S13 = n/2{2a+(n-1)d} = 13/2{2*6+12 *6} = 546

S13-S28 =2436-546 =1890

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.

Hope it's clear.
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Hi All,

While many Test Takers will use a standard, algebraic approach to this question, you can also answer this question by using "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

If you calculate JUST the unit's digit, you'll have the correct answer (since only one answer has a units digit of 8).

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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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\(S13 = 6* (13-1) + 6 = 78\)
\(S28 = 6* (28-1) + 6 = 168\)

Average of evenly spaced \(= \frac{First + Last}{2} = \frac{78 + 168}{2} = \frac{246}{2} = 123\)

Number of terms = Last - first + 1 = 28 - 13 + 1 = 15 + 1 = 16

16 * 123 = 1,968

Answer chocie D

Hello!

COuld someone please clarify to me how do we get to the following?

\(s_1+6(n-1)\)

Kind regards!

Given \(S_n = S_{n-1} + 6\)
Every subsequent term is 6 more than the previous term. So it is an Arithmetic Progression with common difference 6. In an arithmetic progression,

\(T_n = a + (n - 1)*d\)
where a is the first term, \(T_n\) is the nth term and d is the common difference.

So \(S_n = s_1 + (n - 1)*6\)

Calculate the number of terms correctly

S13 ........ S28 will be 16 terms, [a+b-1] , one can calculate S13 = a + 12d and S28 = a + 27d

Here a= 6 and d = 6

After that, this series will be a series of multiple of 6
\(S_n\) = n/2 [first term + last term], Sum of n terms in a series

\(S_{16}\) = 16/2 [78 + 168]

Answer D
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The best and easiest working ever, thanks so much for simplifying

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