 July 10, 2020, 12:57:16 pm

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#### svnflower

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• Respect: 0 « Reply #4455 on: May 15, 2020, 11:37:36 pm »
0
Ahh thanks heaps, this topic is becoming much more clearer to me now I've also tried doing this question a few times but keep getting the wrong answer? Would you mind letting me know where I went downhill?

#### Opengangs

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• Respect: +449 « Reply #4456 on: May 15, 2020, 11:57:59 pm »
+1
Ahh thanks heaps, this topic is becoming much more clearer to me now I've also tried doing this question a few times but keep getting the wrong answer? Would you mind letting me know where I went downhill?

Hi there!
Check your bounds :-)) They should be the points of intersection! To find your points of intersection, set $x^2 = 2x + 3$. This is just a quadratic in $x$ that you can solve! This gives you
\begin{align*}
x^2 - 2x - 3 &= 0 \\
(x - 3)(x + 1) &= 0 \\
x = 3, &\quad x = -1.
\end{align*}

So these are your bounds! The next step is to figure out which curve is the top curve! This can be visualised using a diagram and it's pretty easy to see that $y = 2x + 3$ is the top curve as you successfully worked out! So all we need to do is to evaluate the following integral $\displaystyle \int_{-1}^3 (2x + 3 - x^2)\,dx$, which becomes
\begin{align*}
\int_{-1}^3 (2x + 3 - x^2)\,dx &= \left(x^2 + 3x - \frac{1}{3}x^3\right)\bigg|_{-1}^3 \\
&= \left(9 + 9 - 9\right) - \left(1 - 3 + \frac{1}{3}\right) \\
&= 9 + 2 - \frac{1}{3} \\
&= 11 - \frac{1}{3} \\
&= \frac{32}{3} \\
&= 10 \frac{2}{3}.
\end{align*}
I think the mistake here is really figuring out what the bounds are, which are the points of intersection between the two curves. :-))
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#### svnflower

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• Respect: 0 « Reply #4457 on: May 16, 2020, 01:02:34 am »
+1 Thanks again for all the help today, super helpful!

#### 2020hsc

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• Respect: 0 « Reply #4458 on: June 01, 2020, 03:44:25 pm »
0
hey,

could i get some help on the question below...please and thanks its in the year 12 unit on sequences and series, looking at solving problems with arithmetic sequences and geometric sequences.

'find a and b if a,b,1 forms a GP and b,a,10 forms and AP'

thanks!

#### fun_jirachi

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• Respect: +374 « Reply #4459 on: June 01, 2020, 05:07:32 pm »
+1
Hey there,

Think about what an AP and a GP define for any three consecutive terms. For $n \in \mathbb{Z}^+, n > 2$, the former states that $T_n - T_{n-1} = T_{n+1} - T_n$, while the latter states that $\frac{T_n}{T_{n-1}} = \frac{T_{n+1}}{T_n}$.

Now, considering this, we have that $\frac{b}{a}= \frac{1}{b} \implies a=b^2$. We also have that $a-b = 10-a \implies 2a = b+10$.

See how you go from here « Last Edit: June 01, 2020, 05:36:43 pm by fun_jirachi »
Failing everything, but I'm still Flareon up.

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#### 2020hsc

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• Respect: 0 « Reply #4460 on: June 01, 2020, 05:22:05 pm »
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thanks....but should it not be the other way?

considering the GP is a,b,1 and the AP is b,a,10...then should it not be b/a = 1/b and a-b = 10-a ??

#### fun_jirachi

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• Respect: +374 « Reply #4461 on: June 01, 2020, 05:34:52 pm »
+1
thanks....but should it not be the other way?

considering the GP is a,b,1 and the AP is b,a,10...then should it not be b/a = 1/b and a-b = 10-a ??

Yes, it should! I misread it ~ the point still stands, however - you'll still have a quadratic. I'll edit that now so it's correct. In general, it is handy to know what makes a GP a GP, and what makes an AP an AP, even though these questions are relatively uncommon - the theory is so important to reasoning out further applications Failing everything, but I'm still Flareon up.

HSC 2018: Modern History  | 2U Maths 
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#### 2020hsc

• • Posts: 19
• Respect: 0 « Reply #4462 on: June 01, 2020, 05:52:40 pm »
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Yes, it should! I misread it ~ the point still stands, however - you'll still have a quadratic. I'll edit that now so it's correct. In general, it is handy to know what makes a GP a GP, and what makes an AP an AP, even though these questions are relatively uncommon - the theory is so important to reasoning out further applications ok, sweet. thanks a lot from there then do you solve it as a simultaneous equation or how would be best?

#### Einstein_Reborn_97 « Reply #4463 on: June 01, 2020, 07:15:16 pm »
0
ok, sweet. thanks a lot from there then do you solve it as a simultaneous equation or how would be best?
Yes, solve it using a simultaneous equation. Let me know what answers you get HSC 2020: Advanced English | Advanced Mathematics | Physics | Chemistry | Biology | Studies of Religion 2

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#### 2020hsc

• • Posts: 19
• Respect: 0 « Reply #4464 on: June 01, 2020, 07:27:11 pm »
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Yes, solve it using a simultaneous equation. Let me know what answers you get thanks!!

coming out with b= -2, a = 4 or b= 5/2, a = 25/4

#### Einstein_Reborn_97 « Reply #4465 on: June 01, 2020, 08:00:28 pm »
0
thanks!!

coming out with b= -2, a = 4 or b= 5/2, a = 25/4

Yep, correct. Both solutions work for both the GP and the AP.
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#### RuskiBrah

• • Posts: 6
• Respect: 0 « Reply #4466 on: June 02, 2020, 09:04:26 pm »
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hey guys, was wondering what formula to use for this question and why. cheers

#### Einstein_Reborn_97 « Reply #4467 on: June 02, 2020, 09:30:37 pm »
0
hey guys, was wondering what formula to use for this question and why. cheers
Hey RuskiBrah,

For (a): Use the formula for the sum of a geometric series:
a=45, n=6, r=0.4 (so make sure to use to the correct formula; -1<r<1).
Why? Initial height (first term) is 45 cm. Common ratio, r is 0.4: "grows by 2/5 of its previous growth each month". n in this case is the number of months so is equal to 6.

For (b): Use the formula for limiting sum of an infinite geometric series:
a=45 and r=0.4
Why? The question is asking for the lamb's "final height" and for geometric series where -1<r<1, you can find its limiting sum because the series converge.

Hope that helps! Let me know if you need any further explanations HSC 2020: Advanced English | Advanced Mathematics | Physics | Chemistry | Biology | Studies of Religion 2

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#### RuskiBrah

• • Posts: 6
• Respect: 0 « Reply #4468 on: June 02, 2020, 10:11:00 pm »
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ohh i forgot about that formula. that clears it up, thank you

#### RuskiBrah

• • Posts: 6
• Respect: 0 « Reply #4469 on: June 02, 2020, 10:18:04 pm »
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Hey RuskiBrah,

For (a): Use the formula for the sum of a geometric series:
a=45, n=6, r=0.4 (so make sure to use to the correct formula; -1<r<1).
Why? Initial height (first term) is 45 cm. Common ratio, r is 0.4: "grows by 2/5 of its previous growth each month". n in this case is the number of months so is equal to 6.
Hope that helps! Let me know if you need any further explanations 