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Sequences

by resilient » Wed Apr 09, 2008 5:38 pm
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}?



1,800
1,845
1,890
1,968
2,016


I am having troubles trying to understand these types of questions. I am wondering if anyone can show me where to go to learn ALL about this topic. I almost always get snagged by this and would liek to master it!
Appetite for 700 and I scraped my plate!
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by ritz » Wed Apr 09, 2008 6:13 pm
in the questions pertaining to series, follow this approach...

1. decide what kind of series it is.. (arithmetic, geometric, harmonic etc)
2. Once you are done with this, find out the nth term equation, i.e. try to establish an equation in terms of n (n being the number of term), so that if you put n =1, you get the first term, n = 2, you get the second term. and so on & so forth...
3. once you do that, you can easily solve most of the problems....

Lets take this example now..
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}?

given...
S1 = 6
S2 = 12 & Sn = Sn-1 + 6
As a constant is being added to the previous term, it is an arithmetic series with a (first term) as 6 & d (difference bet terms) as 6.
(step 1 achieved) (step 2 is given, Sn)
so we can make out that
s13 = 6 + (13-1)6 (formula a+(n-1)d)
= 6 + 72 = 78
& similarily
s28 = 6 + (28-1)6
= 6 + 162 = 168
now the set given is
78, 84....168
by applying the sum of arithmetic series formula, {n(a+l)/2}you can easily find out that the answer is...
8*246 = 1968
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nice

by resilient » Wed Apr 09, 2008 8:51 pm
this is great. I sometimes cant see the equation or what its doing. Do you know of a place to get some basic training on this.>?
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by aravind45 » Thu Jul 04, 2013 9:03 am
you can get the formula and it's application from Nova Math Bible or google for Arithmetic progression and Geometric Progression.
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by Deepthi Subbu » Sat Aug 03, 2013 12:56 am
Hi,

The recursive function is Sn =Sn-1 +6. The direct function is Sn= 6n.
So S13 = 6(13) = 78.
S14 = 6(14).................... S28 = 6(28)
So Sum = ( 6*13 + 6*14 + 6*15 + ................ 6*28)
6( 13+14+15+.............. +28)
6( (1+2+3+.....+28)-( 1+2+3+....13 ) ) = 1890 . However thats not the answer. Why so ?
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by Tibin » Tue Dec 17, 2013 8:41 am
@Deepthi Subbu : Silly mistake deepthi. :-)

(1+2+3+.....+28)-( 1+2+3+....13 )= 1+2+3+...12

But what we need instead is 1+2+3+...13

ie,
(1+2+3+.....+28)-(1+2+3+....12 ) = 1+2+3+...13
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