Hi everyone,
I am taking the GMAT in two days and yesterday I started taking a GMAT prep test, which I flunked because it is so much harder than the problems in the 4 GMAT books that I have been studying. I wanted to know if you could help me with some of these questions please!?
Here's a question:
In the arithmetic sequence t(1), t(2), t(3), ..., t(n), the t(1) = 23 and t(n)=t(n-1)-3 for each n > 1. What is the value of n when t(n)=-4?
I already solved it by writing down the numbers -4, -1, 2, 5, 8, 11, 14, 17, 20, 23 and came to the conclusion that if t(1) = 23, then n = 10. But is there a formula that I can use to find the answer? Thanks.
Method 2.
t(n) clearly declines linearly as n increases, so put:
t(n)=kn+c.
When t(1)=23, and t(2)=20, so we have:
k+c=23
2k+c=20,
so subtracting the first from the second gives:
(2k+c)-(k+c)=20-23
or:
k=-3.
Substitute this back into the first equation to ge:
c=26,
so:
t(n)=26-3n,
and the rest is as before.
RonL
Here is a linear algebra approach.
You can solve the recurrence relation.
And initial condition is .
First we solve the homogenous equation,
The charachteristic equation is,
.
Thus, the general solution is,
Where is a constant.
To find a specfic solution to,
We look for a solution in the form,
Substitute,
Thus,
The specific solution is,
Thus, the general solution is the sum of particular and general solution to homogenous equation,
Intitial conditions,
Thus,
And we need to solve,
Interesting. (I learned all my Linear Algebra in a Quantum Physics class, so there are broad gaps in my knowledge.) I couldn't help but notice how similar this solution pattern is to solving a similar looking ODE. Is there any relationship between the two that you know of?
-Dan
Almost everything in differential equations/calculus has its analogue in the
world of difference equations/calculus of finite differences.
Integration by parts has an analogous (two forms actually) summation by
parts formula, which I wasted a lot of time playing with once upon a time.
RonL