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July 16, 2018, 10:38:59 pm

### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 844442 times) Tweet Share

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#### grestal

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9285 on: May 28, 2018, 09:09:24 pm »
+1
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#### PopcornTime

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9286 on: June 01, 2018, 11:06:22 pm »
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Can someone help me with answering this AMC question?

#### noregret

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9287 on: June 06, 2018, 07:22:09 pm »
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Hello, I have three questions about Sequences and Series.

1.How do I solve this question:  The sum of the first six terms of a geometric progression is 63 and the ratio of the fourth term to the sum of the second and third terms is -4. Find the first term and common ratio of the progression.

2. How do I prove that the sum n terms of 1^2+(1^2+2^2)^2+(1^2+2^2+3^2)+(1^2+2^2+3^2+4^2)+............. is 1/12(n)(n+1)^2(n+2)?

3. The sum of the first (n+1) terms of an arithmetic series is 5n^2+13n+8. Find the 1st term, the 10 term and the nth term. I get that answer for the 1st term is S(0+1)=5(0)^2+13(0)+8=8, however I do not understand the answer for the 10th and the nth term. If Un=Sn-Sn-1, then Un+1should equal to Sn-1-Sn-2?

#### TheBigC

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9288 on: June 15, 2018, 12:44:53 pm »
+1
Hello, I have three questions about Sequences and Series.

1.How do I solve this question:  The sum of the first six terms of a geometric progression is 63 and the ratio of the fourth term to the sum of the second and third terms is -4. Find the first term and common ratio of the progression.

2. How do I prove that the sum n terms of 1^2+(1^2+2^2)^2+(1^2+2^2+3^2)+(1^2+2^2+3^2+4^2)+............. is 1/12(n)(n+1)^2(n+2)?

3. The sum of the first (n+1) terms of an arithmetic series is 5n^2+13n+8. Find the 1st term, the 10 term and the nth term. I get that answer for the 1st term is S(0+1)=5(0)^2+13(0)+8=8, however I do not understand the answer for the 10th and the nth term. If Un=Sn-Sn-1, then Un+1should equal to Sn-1-Sn-2?

1.
Therefore, for a geometric series:
${S_{n}}=\frac{a(r^{n}-1)}{r-1}$
$\mbox{(1) } {S_{6}}=\frac{a(r^{6}-1)}{r-1} = 63$
$\mbox{(2) } \frac{S_{4} - S_{3}}{S_{3}-S_{1}} = -4$
$S_{4} - S_{3} = \frac{a(r^{4}-1)}{r-1} - \frac{a(r^{3}-1)}{r-1} = \frac{a(r^{4}-1) - a(r^{3}-1)}{r-1} = \frac{a(r^{4} - r^{3})}{r-1}$
$S_{3} - S_{1} = \frac{a(r^{3}-1)}{r-1} - \frac{a(r^{1}-1)}{r-1} = \frac{a(r^{3}-1) - a(r^{1}-1)}{r-1} = \frac{a(r^{3} - r^{1})}{r-1}$
Thus,
$\frac{S_{4} - S_{3}}{S_{3}-S_{1}} = \frac{\frac{a(r^{4} - r^{3})}{r-1}}{\frac{a(r^{3} - r^{1})}{r-1}} = \frac{a(r^{4} - r^{3})}{a(r^{3} - r^{1})} = \frac{(r^{3} - r^{2})}{(r^{2} - 1)} = \frac{r^{2}(r - 1)}{(r - 1)(r + 1)} = \frac{(r^{2})}{(r + 1)} = -4$
$\frac{(r^{2})}{(r + 1)} = -4 \implies -4(r+1)=r^{2} \implies r^{2} + 4r +4 = (r+2)^{2} = 0$
Thus,
(Common ratio)
$r = -2$
Subbing into (1) gives:
${S_{6}}=\frac{a((-2)^{6}-1)}{(-2)-1} = \frac{a(63)}{-3} = 63$
Therefore,
(First term)
$a = -3$
2. USE MATHEMATICAL INDUCTION OR OTHERWISE.
3.
$S_{n+1} = 5n^{2} + 13n + 8$
For first term:
$S_{0+1} = 5(0)^{2} + 13(0) + 8 = 8$
For 10th term:
$S_{9+1} - S_{8+1} = 5(9)^{2} + 13(9) + 8 - (5(8)^{2} + 13(8) + 8) = 98$
For the nth term:
$S_{(n-1)+1} - S_{(n-2)+1} = 5(n-1)^{2} + 13(n-1) + 8 - (5(n-2)^{2} + 13(n-2) + 8) = 10n-2$
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#### Buoyancy

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9289 on: June 20, 2018, 10:03:09 pm »
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hey guys i need help with finding the area of A2, I can't get my head around it need some explanation.

thanks
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#### TheBigC

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9290 on: June 21, 2018, 04:43:33 pm »
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hey guys i need help with finding the area of A2, I can't get my head around it need some explanation.

thanks

Refer to attachment
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#### Guideme

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9291 on: June 30, 2018, 06:04:22 pm »
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Can anyone answer question 4 pls

:0

#### Hok999

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9292 on: July 01, 2018, 08:27:26 pm »
+1
Can anyone answer question 4 pls

Formula for Euler's Method is: yn+1 = yn + h.f'(xn,yn)
I can link you videos on youtube if you need to understand how this formula came about.

x0=0                                          y0=1
x1= 0+0.1=0.1                           y1= 1 + 0.1 x  (13) = 1.1
x2= 0.1+0.1=0.2                        y2= 1.1 + 0.1 x  (1.13) = 1.2331
x3= 0.2+0.1=0..3                       y3= 1.2331 + 0.1 x (1.23313) = 1.4206
x4= 0.3+0.1=0.4                        y4= 1.4206 + 0.1 x  (1.42063) = 1.7073
x5= 0.4+0.1=0.5                        y5= 1.7073 + 0.1 x  (1.70733) = 2.2049

y is approximately 2.2049

You can also use the spreadsheet function in the CAS/Ti-Inspire to make this process a lot faster. Hope it helps!

#### Mattjbr2

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9293 on: July 04, 2018, 09:56:17 am »
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Could someone please double check for me whether this is correct? How is it possible for c to equal 4?

We know dx/dt=4 when x=0
but x=0 when t=4
hence dx/dt=4 when t=4

So then aren't we supposed to sub in t=4 and dx/dt=4 to get c=44?

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#### RuiAce

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9294 on: July 04, 2018, 09:58:01 am »
+1
Could someone please double check for me whether this is correct? How is it possible for c to equal 4?

We know dx/dt=4 when x=0
but x=0 when t=4
hence dx/dt=4 when t=4

So then aren't we supposed to sub in t=4 and dx/dt=4 to get c=44?

At first glance, I think that $\frac{dx}{dt} = 4$ when $\boxed{x=0}$ has been typo'd. That should've read $t=0$.

Else, yeah I agree the answers are wrong.
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#### Mattjbr2

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9295 on: July 04, 2018, 10:03:32 am »
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At first glance, I think that $\frac{dx}{dt} = 4$ when $\boxed{x=0}$ has been typo'd. That should've read $t=0$.

Else, yeah I agree the answers are wrong.

Both the old and the new (corrected) versions of the worked solutions - as well as the back of the textbook - all agree on the same supposedly incorrect answer!  Haha. The back of the textbook has never been wrong as far as i remember.  That's so strange
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#### RuiAce

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9296 on: July 04, 2018, 10:08:10 am »
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Both the old and the new (corrected) versions of the worked solutions - as well as the back of the textbook - all agree on the same supposedly incorrect answer!  Haha. The back of the textbook has never been wrong as far as i remember.  That's so strange
If the answers have never been wrong then the question is what's wrong.

(But tbh, yeah, the question is wrong here imo. We know that $x=0$ when $t=4$, but that doesn't necessarily mean $x=0$ ONLY when $t=4$. What if $x=0$ when $t=4, 8$?

When you consider that, the question suddenly doesn't become so doable.)
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#### Mattjbr2

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9297 on: July 04, 2018, 10:41:09 am »
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If the answers have never been wrong then the question is what's wrong.

(But tbh, yeah, the question is wrong here imo. We know that $x=0$ when $t=4$, but that doesn't necessarily mean $x=0$ ONLY when $t=4$. What if $x=0$ when $t=4, 8$?

When you consider that, the question suddenly doesn't become so doable.)

Thanks so much!
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#### Yertle the Turtle

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9298 on: July 12, 2018, 03:10:48 am »
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Could someone explain how to reach an equation for this question, I've been struggling with it for a while.
Thanks
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#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9299 on: July 12, 2018, 07:58:12 am »
0
Could someone explain how to reach an equation for this question, I've been struggling with it for a while.
Thanks

Rate of population increase is proportional to current population size can be expressed as

$\frac{dN}{dt}=kN$ where k is some real constant.

Hence, $\frac{dt}{dN}=\frac{k'}{N}$where k' = 1/k.

Then, antidifferentiating with respect to N, we have $t = k'log_e(N) + c$ (from context we can assume N > 0).

At this stage, before solving simultaneously for k' and c, it is convenient to write N in terms of t. This gives

$N=e^{\frac{1}{k'}(t-c)}$

We then use the facts that:
$t = 0, N = N_0$ and $t = 5, N = 2N_0$ to solve simultaneously for k' and c in terms of initial population. This gives:

$k' = \frac{5}{log_e(2)}$ and $c = \frac{-5log_e(N_0)}{log_e(2)}$

And substituting back into our equation for N, we have (with a bit of tidying up):

$N=N_02^{\frac{t}{5}}$

An alternative (much slicker) method is to begin by recognising that any solution to an equation of the form $f'(t)=cf(t)$ will be a function of the form $f(t)=Ae^{bt}$Then substitute in known values (t = 0, f(0) = A, and t = 5, f(5) = 2A) to solve for b in terms of A. This gives $b=\frac{log_e(2)}{5}$This agrees with the above approach, because b is the reciprocal of k'.
« Last Edit: July 12, 2018, 04:47:20 pm by S_R_K »