A123 GMA Questions

[abd123]

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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

[Mao]

This does not require a poll.

Both are arithmetic sequences.

Moderator action: poll removed

[Drunk]

Q1) For a sequence with Tn=15-5n, find the T5 and the sum of the first 25 terms.
Using the formulas and , 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 in the equation is . If we substitute that in, we'll get:

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 .
This is because ! 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:

And that'll eventually boil down to

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 and 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
Sorry mate, but yeah, there you go.

[gossamer]

A geometric series question.

Q.

A.

I don't think the answer is right: it should be

Use the formula for the sum of a (finite) geometric series:

[Truck]

A geometric series question.

Q.

A.

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.

[brightsky]

'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)

[brightsky]

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

[brightsky]

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

[xZero]

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.

[gossamer]

Trig simplication question.
1.
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.