Sequences

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

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.>?

you can get the formula and it's application from Nova Math Bible or google for Arithmetic progression and Geometric Progression.

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 ?

@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