What is the 25th term of this sequence?

[pujaverma]

*this is from a gmat flashcard

Sn = Sn-1 - 10 and S3 = 0

I calculated the answer the long way. This was easy, since this was only the 25th term. But I don't understand how the equation to solve this is derived, and I want to understand it in case the test asks for the 500th term. The explanation notes that each term is 10 less than the previous one, so Sn = -10n + k, where is k is a constant (how do you get this equation?)

[Ben Ku]

All you need to do to obtain a formula is to observe the pattern of the sequence.

If S3 = 0, then S4 = -10, S5 = -20, etc.

We can go the other way and get S1 = 20, S2 = 10.

So basically, if you want S7, you have to subtract 10 six times from 20 [S7 = 20 - 6(10) = 20 - 60 = -40]
If you want S16, you have to subtract 10 fifteen times from 20 [S16 = 20 - 15(10) = 20 - 150 = -130]
If you want Sn, you have to subtract 10 n-1 times from 20 [Sn = 20 - (n-1)(10)].

This is one way to find a formula for a sequence.

*this is from a gmat flashcard

Sn = Sn-1 - 10 and S3 = 0

I calculated the answer the long way. This was easy, since this was only the 25th term. But I don't understand how the equation to solve this is derived, and I want to understand it in case the test asks for the 500th term. The explanation notes that each term is 10 less than the previous one, so Sn = -10n + k, where is k is a constant (how do you get this equation?)

[thatsku]

why couldn't you have just gone with Sn = (n-3)*-10 rather than deriving S1 and S2 terms?

[nelloriumrao]

Correct me if i am wrong.

given S3=0 & Sn = Sn-1 - 10
So we can find S2 = 10, S1 = 10 using recursive form of Eq

From direct Eq., which is Sn = n (K) + X ; K = common diff & X = constant
S3 = 3.K + X;
but we know k=-10 (S2-S1);
So, Sn = n(-10)+k; in here substitute value of S3=0 (given)

0=3(-10)+X => X=30
Sn=n(10)+30
S25= 25*(-10)+30 => -220

[RonPurewal]

why couldn't you have just gone with Sn = (n-3)*-10 rather than deriving S1 and S2 terms?

well, yeah, that's correct. but ... where'd you get it?

it's true that "solutions by inspection" are often possible for problems like this one, but posting them here isn't going to help anybody.

--

in general, the equations for these sorts of series are indistinguishable from the equations for straight lines.

in this case, you can take any two points - such as (3, 0) and (4, -10) - and derive the equation y = mx + b, exactly as you would in a coordinate system.

always look for connections between different solution techniques!