# GMAT questions

• Jan 18th 2007, 09:15 AM
crazirani
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.
• Jan 18th 2007, 09:44 AM
CaptainBlack
Quote:

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
• Jan 18th 2007, 10:00 AM
CaptainBlack
Quote:

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
• Jan 18th 2007, 10:16 AM
ThePerfectHacker
Quote:

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,
$-3n+26=-4$
$-3n=-30$
$n=-10$
• Jan 18th 2007, 10:58 AM
crazirani
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.
• Jan 18th 2007, 11:07 AM
crazirani
Thanks TPH.
• Jan 18th 2007, 12:24 PM
topsquark
Quote:

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
• Jan 18th 2007, 01:33 PM
ThePerfectHacker
Quote:

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.
• Jan 18th 2007, 02:21 PM
CaptainBlack
Quote:

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