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If S is the infinite sequence S1 = 6, S2 = 12, ..., Sn =Sn-1 [#permalink]
 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]
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]
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]
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]
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]
06 Oct 2010, 05:17
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are we supposed to know this formula for GMAT?
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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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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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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]
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]
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]
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]
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]
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: s in infinite sequence
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15 Feb 2012, 11:19
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