1. ## GMAT questions

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.

2. Originally Posted by crazirani
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.
t(n)=t(n-1)-3=t(n-2)-3-3=...=t(n-(n-1))-3(n-1),

but t(n-(n-1))=t(1), so:

t(n)=t(1)-3(n-1),

so now we have:

t(n)=23-3n+3=26-3n

If t(n)=4, we have:

-4=26-3n,

or:

3n=26+4=30,

so n=10.

RonL

3. Originally Posted by crazirani
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

4. Originally Posted by crazirani
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?
Here is a linear algebra approach.

You can solve the recurrence relation.

$t(n)-t(n-1)=-3$
And initial condition is $t(1)=23$.

First we solve the homogenous equation,
$t(n)-t(n-1)=0$
The charachteristic equation is,
$k-1=0$
$k=1$.
Thus, the general solution is,
$t(n)=C(1)^n=C$
Where $C$ is a constant.

To find a specfic solution to,
$t(n)-t(n-1)=-3$
We look for a solution in the form,
$t(n)=an$
$t(n-1)=a(n-1)=an-a$
Substitute,
$an-an+a=-3$
$a=-3$
Thus,
The specific solution is,
$t(n)=-3n$
Thus, the general solution is the sum of particular and general solution to homogenous equation,
$t(n)=-3n+C$
Intitial conditions,
$t(1)=-3(1)+C=+23$
$C=26$
Thus,
$t(n)=-3n+26$

And we need to solve,
$-3n+26=-4$
$-3n=-30$
$n=-10$

5. Thanks Ron, I'll have to study your responses very carefully and try to do some more problems to really understand what's going on.

6. Thanks TPH.

7. Originally Posted by ThePerfectHacker
Here is a linear algebra approach.

You can solve the recurrence relation.

$t(n)-t(n-1)=-3$
And initial condition is $t(1)=23$.

First we solve the homogenous equation,
$t(n)-t(n-1)=0$
The charachteristic equation is,
$k-1=0$
$k=1$.
Thus, the general solution is,
$t(n)=C(1)^n=C$
Where $C$ is a constant.

To find a specfic solution to,
$t(n)-t(n-1)=-3$
We look for a solution in the form,
$t(n)=an$
$t(n-1)=a(n-1)=an-a$
Substitute,
$an-an+a=-3$
$a=-3$
Thus,
The specific solution is,
$t(n)=-3n$
Thus, the general solution is the sum of particular and general solution to homogenous equation,
$t(n)=-3n+C$
Intitial conditions,
$t(1)=-3(1)+C=+23$
$C=26$
Thus,
$t(n)=-3n+26$

And we need to solve,
$-3n+26=-4$
$-3n=-30$
$n=-10$
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

8. Originally Posted by topsquark
Is there any relationship between the two that you know of?
Eigenvalues.
The chrachteristic polynomial that we get from,
$\det (k\bold{I}-\bold{A})=0$.

If,
$A\bold{x}=\bold{b}$
Is a linear system.
With $n$ equations and $m$ unknowns.

And then all solutions are,
$x_0+\mbox{Nullspace}(A)$
Where $\mbox{Nullspace}(A)$ is the basis for $A\bold{x}=\bold{0}$ meaning "the general solution".

See the same pattern is here as well.

9. Originally Posted by topsquark
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?
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