ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: abd123 on April 12, 2011, 12:02:44 pm

Title: A123 GMA Questions
Post by: abd123 on April 12, 2011, 12:02:44 pm: Help VNERS :)

Arithmetic Questions

Q1) For a sequence with Tn=15-5n, find the T5 and the sum of the first 25 terms.

Q2) The sum of the first n terms of a particular sequence is given by Sn=17n-3n^2
   a) Find the expression for the sum to (n-1) terms.
   b) Find an expression for the nth term of the sequence
   c) Show that the corresponding sequence is arithmetic and find a and d.

Thanks :)
Title: Re: Sequence and Series
Post by: Mao on April 12, 2011, 12:18:29 pm: This does not require a poll.

Both are arithmetic sequences.

Moderator action: poll removed
Title: Re: Sequence and Series
Post by: Drunk on April 16, 2011, 08:42:57 pm: Q1) For a sequence with Tn=15-5n, find the T5 and the sum of the first 25 terms.
Using the formulas $t_{n}=a+(n-1)d$ and $S_{n}=\frac{n}{2}[2a+(n-1)d]$ , you can pretty easily work that out, so I'm not going to go in depth there.

Q2) The sum of the first n terms of a particular sequence is given by Sn=17n-3n^2
a) Find the expression for the sum to (n-1) terms.
So in this case, the $n$ in the equation is $(n-1)$ . If we substitute that in, we'll get:
$S_{n-1}=17(n-1)-3(n-1)^{2}$
So, you can expand that yourself because I'm too lazy to type it out LOL

b) Find an expression for the nth term of the sequence
Okay, well because of the two equations that the question's given us and the one we just worked out, we can work out $t_{n}$ .
This is because $t_{n}=S_{n}-S_{n-1}$ ! I'd explain that to you, but yeah, I don't wanna waste too much time. You can ask me again if you want me to and I'll explain it in another post. Anyway, from this, you'll get:
$t_{n}=17n-3n^{2}-[17(n-1)-3(n-1)^{2}]$
And that'll eventually boil down to $t_{n}=20-6n$

c) Show that the corresponding sequence is arithmetic and find a and d.
I've got no clue as to how to properly answer this question, but I'll guess that what it's asking for is something like the first three terms?
Anyway, they're going to be 14, 8 and 2 (from the equation above). So yeah, as you can see, it's clearly going down by 6, so that'll be your $d$ and $t_{1}$ is obviously 14 there. That does answer the question, but pretty informally.

AND yeah, I actually can't help you with that last one there seeing as I've been struggling with it too :P
Sorry mate, but yeah, there you go.
Title: Re: Sequence and Series
Post by: gossamer on April 17, 2011, 12:47:14 am: Quote from: abd123 on April 16, 2011, 05:29:41 pm
A geometric series question.

Q. $1-x^2+x^4-x^6+...+x^2^m (m/is/even)$

A. $x^2^m+1/x^2+1$
I don't think the answer is right: it should be $\frac{x^2^m^+^2+1}{x^2+1}$

Use the formula for the sum of a (finite) geometric series: $S_{n}=\frac{a(1-r^n)}{1-r}$
Title: Re: Sequence and Series
Post by: Truck on April 17, 2011, 03:20:10 am: Quote from: abd123 on April 16, 2011, 05:29:41 pm
A geometric series question.

Q. $1-x^2+x^4-x^6+...+x^2^m (m/is/even)$

A. $x^2^m+1/x^2+1$

I've seen two ways to do this question, one my tutor did which I barely understand and one my maths teacher taught which makes a bit more sense to me, so I'll post the second one: (I'm a noob with LaTeX, SORRY I'll try to make this readable anyway.

We know that a=1, r=-x^2, tn = x^2m. Therefore we need to find n and Sn.

tn = 1 x (-x^2)^n-1 = x^2m
tn = (-x^2)^n-1 = x^2m

We can classify {-x^2+x^4-x^6+...+x^2^m} as "m" terms, and say that when added it equals Sm-1 (you can't forget the 1 on the outside).

Therefore Sm = a(r^m - 1)/r-1
= -x^2(x^2m - 1)/-x^2 -1
= - x^2(x^2m + 1)(x^2 + 1).

Not forgetting the -1, Sm = 1 + x^2(x^2m + 1)/x^2 + 1
= x^2 + 1 + x^2m+2 - x^2/x^2 +1
= x^2m+2 + 1/x^2 + 1

Hopefully that made sense =). I will learn LaTeX soon, but it's 3:20am right now so not the right time LOL.
Title: Re: Sequence and Series
Post by: brightsky on April 17, 2011, 09:05:08 pm: 'Another' way to look at it:
Let S = 1 - x^2 + x^4 - x^6 + ... + x^(2m)
then Sx^2 = x^2 - x^4 + x^6 - ... - x^(2m) + x^(2m + 2)
So then Sx^2 + S = 1 - x^2 + x^4 - x^6 + ... + x^(2m) + x^2 - x^4 + x^6 - ... - x^(2m) + x^(2m + 2)
S(x^2 + 1) = 1 + x^(2m+2)
S = (1 + x^(2m+2))/(x^2+1)
Title: Re: A123 GMA Questions
Post by: brightsky on April 17, 2011, 11:00:51 pm: Well the original triangle would have perimeter 3p. Then the next one would have perimeter 1/2(3p). And then the next one 1/2^2 (3p) and so on. So the nth one would have 1/2^(n-1) (3p). The sum of all the perimeters would be:
S = 3p + 1/2(3p) + 1/2^2 (3p) + ...
S*(1/2) = 1/2(3p) + 1/(2^2)(3p) + ...
Hence S - 1/2*S = 3p
1/2 S = 3p
S = 6p.
(Note that the series converges so this equation is allowed.)

Area of the nth triangle is given by:
p/2^(n) * sqrt((p/2^(n-1))^2 - (p/2^n)^2)
= p/2^(n) * sqrt(3)p/2^n
So the sum is: S = sqrt(3)p^2/2 + sqrt(3)p^2/4 + sqrt(3)p^2/8 + ...
1/2S = sqrt(3)p^2/4 + sqrt(3)p^2/8 + ...
S - 1/2 S = sqrt(3)p^2/2
S = sqrt(3)p^2
Title: Re: A123 GMA Questions
Post by: brightsky on May 08, 2011, 12:24:48 pm: 4/(x-1)^2(2x + 1) = A/(2x + 1) + B/(x-1) + C/(x-1)^2
4 = A(x-1)^2 + B(x-1)(2x+1) + C(2x+1) = A(x^2 - 2x + 1) + B(2x^2 - x - 1) + C(2x+1) = Ax^2 - 2Ax + A + 2Bx^2 - Bx - B + 2Cx + C = x^2(A + 2B) - x(2A + B - 2C) + (A - B + C)
So A + 2B = 0...(1)
2A + B - 2C = 0...(2)
A - B + C = 4...(3)
(1)*2: 2A + 4B = 0..(4)
(4)-(2): 3B + 2C = 0..(5)
(3)*2: 2A - 2B + 2C = 8..(6)
(6)-(2): -3B + 4C = 8..(7)
(5)+(7): 6C = 8, C = 4/3
so B = -8/9, A = 16/9 by substitution
Title: Re: A123 GMA Questions
Post by: xZero on June 19, 2011, 12:52:17 pm: let a = first number and b = second number, p=a+b. we know that a is proportional to x while b is proportional to y^2 => a=mx, b=ny^2 where m and n are arbitrary constant. Using the condition 14 = m(1)+n(2)^2 and 31=m(2)+n(3)^2, solve for m and n then you can find p when x=3 and y=4.
Title: Re: A123 GMA Questions
Post by: gossamer on August 07, 2011, 04:54:50 pm: Quote from: abd123 on August 07, 2011, 04:45:01 pm
Trig simplication question.
1. $\frac{sin^4\Theta -cos^4\Theta }{sin\Theta +cos\Theta }\cdot\frac{cos\Theta }{sincos\Theta -cos^2\Theta }$
Answer: 1

Is the denominator of the second fraction sin(theta)cos(theta)?

Try and see where you can use factorising and cancelling, difference of perfect squares, and the pythagorean identity sin^2(theta) + cos^2(theta) = 1 to simplify it. :)