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

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

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New post 04 Oct 2010, 07:17
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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
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Re: s in infinite sequence  [#permalink]

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New post 04 Oct 2010, 07:51
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anilnandyala 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


Given: \(s_1=6\) and \(s_n=s_{n-1}+6=s_1+6(n-1)\).

Question: sum of 16 elements from this sequence \(s_{13}+s_{14}+...+s_{28}=?\)

As \(s_n=s_1+6(n-1)\) then \(s_{13}=6+6(13-1)=78\) and \(s_{28}=6+6(28-1)=168\).

Sum of 16 evenly spaced terms would be \(\frac{first \ term+last \ term}{2}*# \ of \ terms=\frac{s_{13}+s_{28}}{2}*16=\frac{78+168}{2}*16=1968\).

Answer: D.
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Re: s in infinite sequence  [#permalink]

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New post 04 Oct 2010, 11:50
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
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Re: s in infinite sequence  [#permalink]

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New post 04 Oct 2010, 12:38
Sn = 6*n...
S1 = 6 * 1; S2 = 6*2; ...

From S13 to S28: 6*13 + 6*14 + ... + 6*28 = 6* (13 + 14 + ... + 28)

13 + 14 + ...= 16 * (13+28)/2 = 328

Therefore 6 * 328 = 1968 (only term that ends with 8)
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Re: s in infinite sequence  [#permalink]

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New post 04 Oct 2010, 17:49
s1=6, s2=12.
so s13=13*6 and s28=28*6
so the sum = number of terms*(first term + last term)/2
= 16*6*(13+28)/2
=1968
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Re: s in infinite sequence  [#permalink]

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New post 06 Oct 2010, 05:17
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are we supposed to know this formula for GMAT?
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Sequence  [#permalink]

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New post 13 Nov 2010, 17:59
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
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Re: Sequence  [#permalink]

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New post 13 Nov 2010, 18:59
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MateoLibre wrote:
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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Re: Sequence  [#permalink]

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New post 13 Nov 2010, 22:25
need to add 6x13 + 6x14 +....6x28 = 6(13 + 14 + ...28)
so 6 * (13+28)/2 * 16 = 3 * 41 * 16 = 1968

(sum of an arithmatic sequence = (first term + last term)/2 * no of terms)
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Re: s in infinite sequence  [#permalink]

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New post 14 Nov 2010, 04:26
prashantbacchewar wrote:
are we supposed to know this formula for GMAT?


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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Re: s in infinite sequence  [#permalink]

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New post 25 Mar 2011, 04:40
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
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Re: s in infinite sequence  [#permalink]

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New post 27 Mar 2011, 01:39
(2)^1/3, (5)^1/6, (10)^1/10, (30)^1/15

s13 = s1 + (12) * 6

=> s13 = 13 * 6 = 78

s28 = s1 + 27 * 6

s28 = 28 * 6 = 168

So Sum = 16 * (78 + 168)/2

= 8 * 246

The answer must end with last digit as 8 and We can stop multiplying here as there is only 1 answer like that.

= 1968

Answer - D
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Re: s in infinite sequence  [#permalink]

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New post 15 Feb 2012, 06:50
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....
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Re: s in infinite sequence  [#permalink]

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New post 15 Feb 2012, 11:19
prakarp wrote:
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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Re: If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = Sn-  [#permalink]

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New post 29 May 2017, 10:24
anilnandyala 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


Sn = 6*N

S13 = 6*13

S28 = 6*28

# of terms = 28 -13 + 1 = 16

(6*13+6*28)/2

= 3*(13+28)

= 3*41 = 123

Once we have the mean of the set, we can multiply it by the number of terms in the set to arrive at the sum of those terms.

123*16 = 1968
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Re: If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = Sn-  [#permalink]

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New post 15 Mar 2018, 12:55
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).

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

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New post 30 Mar 2018, 10:58
anilnandyala 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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Re: If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = Sn-  [#permalink]

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New post 05 Oct 2018, 15:39
anilnandyala 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


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

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
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Re: If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn = Sn- &nbs [#permalink] 05 Oct 2018, 15:39
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