# Thread: HSC 2017-2018 Maths Marathon

1. ## Re: HSC 2017 Maths (Advanced) Marathon

To add onto the marathon, here's a new question:

Capture.PNG
$The\quad diagram\quad shows\quad the\quad graphs\quad of\quad y\quad =\quad \sqrt { 3 } cos\quad x\quad and\quad y\quad =\sin { \quad x } .\quad The\quad first\quad two\quad points\quad of\quad intersection\quad to\quad the\quad right\quad of\quad the\quad y-axis\quad are\quad labelled\quad A\quad and\quad B.\quad \\ i)\quad Solve\quad the\quad equation\quad \sqrt { 3 } cos\quad x\quad =\quad sin\quad x\quad to\quad find\quad the\quad coordinates\quad of\quad A\quad and\quad B.\\ ii)\quad Find\quad the\quad shaded\quad region\quad in\quad the\quad diagram.$

2. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by athena13
To add onto the marathon, here's a new question:

Capture.PNG
$The\quad diagram\quad shows\quad the\quad graphs\quad of\quad y\quad =\quad \sqrt { 3 } cos\quad x\quad and\quad y\quad =\sin { \quad x } .\quad The\quad first\quad two\quad points\quad of\quad intersection\quad to\quad the\quad right\quad of\quad the\quad y-axis\quad are\quad labelled\quad A\quad and\quad B.\quad \\ i)\quad Solve\quad the\quad equation\quad \sqrt { 3 } cos\quad x\quad =\quad sin\quad x\quad to\quad find\quad the\quad coordinates\quad of\quad A\quad and\quad B.\\ ii)\quad Find\quad the\quad shaded\quad region\quad in\quad the\quad diagram.$
http://imgur.com/YqJJwLA

3. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by athena13
To add onto the marathon, here's a new question:

Capture.PNG
$The\quad diagram\quad shows\quad the\quad graphs\quad of\quad y\quad =\quad \sqrt { 3 } cos\quad x\quad and\quad y\quad =\sin { \quad x } .\quad The\quad first\quad two\quad points\quad of\quad intersection\quad to\quad the\quad right\quad of\quad the\quad y-axis\quad are\quad labelled\quad A\quad and\quad B.\quad \\ i)\quad Solve\quad the\quad equation\quad \sqrt { 3 } cos\quad x\quad =\quad sin\quad x\quad to\quad find\quad the\quad coordinates\quad of\quad A\quad and\quad B.\\ ii)\quad Find\quad the\quad shaded\quad region\quad in\quad the\quad diagram.$
Oh my god the number of times \quad is used in that TEX code... anyways, is that picture from the MIF textbook?

4. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by InteGrand
$\noindent Given n real numbers x_{1},x_{2},\ldots, x_{n} (n a positive integer), a \textsl{weighted average} of these is a combination of them of the form \sum_{i=1}^{n}\alpha_{i}x_{i}, where the \alpha_{i} are non-negative numbers that sum to 1. E.g. \frac{1}{3}x_{1} + \frac{2}{3}x_{2} is an example of a weighted average of \left\{x_{1},x_{2}\right\}.$

$\noindent Let \overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K} each be weighted averages of \left\{x_{1},x_{2},\ldots, x_{n}\right\} (where K is a positive integer). Show that any weighted average of \left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\} is a weighted average of \left\{x_{1},x_{2},\ldots, x_{n}\right\}.$
$\noindent Now, the weighted average of \left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\} is given by \\\\ \sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i}=(\alpha')_1\overline{x}_{1} + (\alpha')_2\overline{x}_{2} + ... + (\alpha')_K\overline{x}_{K}\\\\But each \overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K} is a weighted average of \left\{x_{1},x_{2},\ldots, x_{n}\right\}. Thus, let us substitute \sum_{i=1}^{n}\alpha_ix_i for \overline{x}_1,\overline{x}_2, ... , \overline{x}_K\\\\ Thus \\\\ (\alpha')_1\overline{x}_{1} + (\alpha')_2\overline{x}_{2} + ... + (\alpha')_K\overline{x}_{K} \\ = (\alpha')_1\sum_{i=1}^{n}\alpha_ix_i + (\alpha')_2\sum_{i=1}^{n}\alpha_ix_i + ... + (\alpha')_3\sum_{i=1}^{n}\alpha_ix_i \\\\ We factorise \sum_{i=1}^{n}\alpha_ix_i and now we have \\\\ \sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i}=\sum_{i=1}^{n}\alpha_ix_i \left\{(\alpha')_1 + (\alpha')_2 + ... + (\alpha')_K \right\}$

$\noindent But we know that the sum of these (\alpha')_K, where K \in \mathbb{Z}^+, is equal to 1 (by given data). And thus, \\\\ \sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i} = \sum_{i=1}^{n}\alpha_ix_i \\\\ This means that the weighted average of \left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\} is a weighted average of \left\{x_{1},x_{2},\ldots, x_{n}\right\}.$

5. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by 1729
Excuse my notation, but here we go (correct me if I am wrong InteGrand)

$\noindent Now, the weighted average of \left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\} is given by \\\\ \sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i}=(\alpha')_1\overline{x}_{1} + (\alpha')_2\overline{x}_{2} + ... + (\alpha')_K\overline{x}_{K}\\\\But each \overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K} is a weighted average of \left\{x_{1},x_{2},\ldots, x_{n}\right\}. Thus, let us substitute \sum_{i=1}^{n}\alpha_ix_i for \overline{x}_1,\overline{x}_2, ... , \overline{x}_K\\\\ Thus \\\\ (\alpha')_1\overline{x}_{1} + (\alpha')_2\overline{x}_{2} + ... + (\alpha')_K\overline{x}_{K} \\ = (\alpha')_1\sum_{i=1}^{n}\alpha_ix_i + (\alpha')_2\sum_{i=1}^{n}\alpha_ix_i + ... + (\alpha')_3\sum_{i=1}^{n}\alpha_ix_i \\\\ We factorise \sum_{i=1}^{n}\alpha_ix_i and now we have \\\\ \sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i}=\sum_{i=1}^{n}\alpha_ix_i \left\{(\alpha')_1 + (\alpha')_2 + ... + (\alpha')_K \right\}$

$\noindent But we know that the sum of these (\alpha')_K is equal to 1 (by given data). And thus, \\\\ \sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i} = \sum_{i=1}^{n}\alpha_ix_i \\\\ In English, this means that the weighted average of \left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\} is a weighted average of \left\{x_{1},x_{2},\ldots, x_{n}\right\}, as required.$
$\noindent Well what you've done here is use the same weighted average for each \overline{x}_{j} (j=1,2,\ldots , K). This is because you let them all be equal to \sum_{i =1}^{n}\alpha_{i}x_{i} (so you gave each of them necessarily the same set of weights \alpha_{i}). But the question was actually to prove it for the \overline{x}_{j} being any arbitrary weighted averages, not necessarily all the same (so each of the K weighted averages has its own set of \alpha weights).$

6. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by InteGrand
$\noindent Well what you've done here is use the same weighted average for each \overline{x}_{j} (j=1,2,\ldots K). This is because you let them all be equal to \sum_{i =1}^{n}\alpha_{i}x_{i} (so you gave each of them necessarily the same set of weights \alpha_{i}). But the question was actually to prove it for the \overline{x}_{j} being any arbitrary weighted averages, not necessarily all the same (so each of the K weighted averages has its own set of \alpha weights).$
\noindent \overline{x}_1 = \alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n\\ \overline{x}_2 = \beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n \\ \ldots \\ \overline{x}_K = \gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n \\\\ Now the weighted average of \left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\} is given by \\ \begin{align*} \sum_{i=1}^K \delta_i\overline{x}_i &= \delta_1\overline{x}_1 + \delta_2\overline{x}_2 + ... + \delta_K\overline{x}_K \\ &= \delta_1(\alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n) + \delta_2(\beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n) + ... + \delta_K(\gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n) \\ &= (\delta_1\alpha_1+\delta_2\beta_1+ ... + \delta_K\gamma_1)x_1 + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2)x_2 + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n)x_n \end{align*}

\noindent For this to be a weighted average of \left\{x_{1},x_{2},\ldots, x_{n}\right\} then \\ (\delta_1\alpha_1+\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) must equal 1. Let us test. Now,\\ \begin{align*}(\delta_1\alpha_1+&\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) \\ &= \delta_1(\alpha_1 + \alpha_2 + ... + \alpha_n) + \delta_2(\beta_1+\beta_2+...+\beta_n) + ... + \delta_K(\gamma_1+\gamma_2+...+\gamma_n) \end{align*} \\ Now, we know that the the sum of the weights \alpha, \beta and \gamma is equal to 1.

\noindent Thus, we now have \\\\\delta_1(\alpha_1 + \alpha_2 + ... + \alpha_n) + \delta_2(\beta_1+\beta_2+...+\beta_n) + ... + \delta_K(\gamma_1+\gamma_2+...+\gamma_n) \\ = \delta_1+\delta_2 + ... + \delta_K. \\\\ The sum of these weights is also equal to 1, therefore we can conclude that \\ \begin{align*} (\delta_1\alpha_1+&\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) = 1 \end{align*} \\ Since this is equal to one, we can conclude that any weighted average of \left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\} is a weighted average of \left\{x_{1},x_{2},\ldots, x_{n}\right\}, given the provided condition.

7. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by 1729
Okay, I am going to use different greek letters for different weights, although not sure if this is reasonable notation. Now correct me if this is wrong.. although I am fairly confident that I got it.. this took ages to type my apologies

\noindent \overline{x}_1 = \alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n\\ \overline{x}_2 = \beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n \\ \ldots \\ \overline{x}_K = \gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n \\\\ Now the weighted average of \left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\} is given by \\ \begin{align*} \sum_{i=1}^K \delta_i\overline{x}_i &= \delta_1\overline{x}_1 + \delta_2\overline{x}_2 + ... + \delta_K\overline{x}_K \\ &= \delta_1(\alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n) + \delta_2(\beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n) + ... + \delta_K(\gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n) \\ &= (\delta_1\alpha_1+\delta_2\beta_1+ ... + \delta_K\gamma_1)x_1 + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2)x_2 + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n)x_n \end{align*}

\noindent For this to be a weighted average of \left\{x_{1},x_{2},\ldots, x_{n}\right\} then \\ (\delta_1\alpha_1+\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) must equal 1. Let us test. Now,\\ \begin{align*}(\delta_1\alpha_1+&\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) \\ &= \delta_1(\alpha_1 + \alpha_2 + ... + \alpha_n) + \delta_2(\beta_1+\beta_2+...+\beta_n) + ... + \delta_K(\gamma_1+\gamma_2+...+\gamma_n) \end{align*} \\ Now, we know that the the sum of the weights \alpha, \beta and \gamma is equal to 1.

\noindent Thus, we now have \\\\\delta_1(\alpha_1 + \alpha_2 + ... + \alpha_n) + \delta_2(\beta_1+\beta_2+...+\beta_n) + ... + \delta_K(\gamma_1+\gamma_2+...+\gamma_n) \\ = \delta_1+\delta_2 + ... + \delta_K. \\\\ The sum of these weights is also equal to 1, therefore we can conclude that \\ \begin{align*} (\delta_1\alpha_1+&\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) = 1 \end{align*} \\ Since this is equal to one, we can conclude that any weighted average of \left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\} is a weighted average of \left\{x_{1},x_{2},\ldots, x_{n}\right\}.
Well done.

8. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by 1729
Okay, I am going to use different greek letters for different weights, although not sure if this is reasonable notation. Now correct me if this is wrong.. although I am fairly confident that I got it.. this took ages to type my apologies

\noindent \overline{x}_1 = \alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n\\ \overline{x}_2 = \beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n \\ \ldots \\ \overline{x}_K = \gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n \\\\ Now the weighted average of \left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\} is given by \\ \begin{align*} \sum_{i=1}^K \delta_i\overline{x}_i &= \delta_1\overline{x}_1 + \delta_2\overline{x}_2 + ... + \delta_K\overline{x}_K \\ &= \delta_1(\alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n) + \delta_2(\beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n) + ... + \delta_K(\gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n) \\ &= (\delta_1\alpha_1+\delta_2\beta_1+ ... + \delta_K\gamma_1)x_1 + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2)x_2 + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n)x_n \end{align*}

\noindent For this to be a weighted average of \left\{x_{1},x_{2},\ldots, x_{n}\right\} then \\ (\delta_1\alpha_1+\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) must equal 1. Let us test. Now,\\ \begin{align*}(\delta_1\alpha_1+&\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) \\ &= \delta_1(\alpha_1 + \alpha_2 + ... + \alpha_n) + \delta_2(\beta_1+\beta_2+...+\beta_n) + ... + \delta_K(\gamma_1+\gamma_2+...+\gamma_n) \end{align*} \\ Now, we know that the the sum of the weights \alpha, \beta and \gamma is equal to 1.

\noindent Thus, we now have \\\\\delta_1(\alpha_1 + \alpha_2 + ... + \alpha_n) + \delta_2(\beta_1+\beta_2+...+\beta_n) + ... + \delta_K(\gamma_1+\gamma_2+...+\gamma_n) \\ = \delta_1+\delta_2 + ... + \delta_K. \\\\ The sum of these weights is also equal to 1, therefore we can conclude that \\ \begin{align*} (\delta_1\alpha_1+&\delta_2\beta_1+ ... + \delta_K\gamma_1) + (\delta_1\alpha_2+\delta_2\beta_2+ ... + \delta_K\gamma_2) + ... + (\delta_1\alpha_n+\delta_2\beta_n+...+ \delta_K\gamma_n) = 1 \end{align*} \\ Since this is equal to one, we can conclude that any weighted average of \left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\} is a weighted average of \left\{x_{1},x_{2},\ldots, x_{n}\right\}.
This is HSC maths??

9. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by boredofstudiesuser1
This is HSC maths??
It can be done using only high school algebra as 1729 did, but admittedly it probably wouldn't come up in a HSC paper (not without any guidance anway).

You can find out more about weighted averages at the Wikipedia article for it: https://en.wikipedia.org/wiki/Weighted_arithmetic_mean .

10. ## Re: HSC 2017 Maths (Advanced) Marathon

I came across a question in the Cambridge textbook today and I haven't looked at the answer yet.

$\int^{5}_{4} \ dx=$

I'm assuming its a constant of 1 next to the dx when we integrate it? Somebody could confirm

11. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by davidgoes4wce
I came across a question in the Cambridge textbook today and I haven't looked at the answer yet.

$\int^{5}_{4} \ dx=$

I'm assuming its a constant of 1 next to the dx when we integrate it? Somebody could confirm
Yes, that's right (it's common practice to drop the '1' and write it like they've written it).

12. ## Re: HSC 2017 Maths (Advanced) Marathon

Congruency test

I came across this question tonight and wanted to clear a couple of things up. At times could we use either RHS test or SAS test to explain parts (b) and (f)? They are exactly the same shape as one can see and the Cambridge textbook gives 2 solutions:

The answer for part (b) was RHS in the textbook

The answer for part (f) was SAS in the text book

13. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by davidgoes4wce
Congruency test

I came across this question tonight and wanted to clear a couple of things up. At times could we use either RHS test or SAS test to explain parts (b) and (d)? They are exactly the same shape as one can see and the Cambridge textbook gives 2 solutions:

The answer for part (b) was RHS in the textbook

The answer for part (f) was SAS in the text book
I would say no (may be wrong).

In (b), the angle is not included within the two sides marked and therefore can only be RHS (SAS is side, included angle, side)

On the other hand, in (f) it is SAS as they are giving you two sides and an included angle but the hypotenuse is not one of the sides. Therefore it is directly SAS and I would say, indirectly RHS.

14. ## Re: HSC 2017 Maths (Advanced) Marathon

^^ I can see it now . It's something that I never really looked into specifically.

SO I take it, if they give the hypotenuse length and a right angle, we can conclude its a RHS . (also one of the adjacent or opposite lengths)

If we are given a right angle, adjacent and an opposite length, then its a SAS.

15. ## Re: HSC 2017 Maths (Advanced) Marathon

I was also looking to compare (b) v (f)....................earlier I mentioned (b) v (d)

16. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by davidgoes4wce
I was also looking to compare (b) v (f)....................earlier I mentioned (b) v (d)
Yeah, I just assumed so since d had no angle marked or hypotenuse

17. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by boredofstudiesuser1
Yeah, I just assumed so since d had no angle marked or hypotenuse
I learnt something new I guess, I always had the assumption that with congruency tests, that if there was a right angle in there, it was always RHS.

Not sure if the majority of HSC and prelim students knew that in the back of their mind. (And I'm not afraid to say this was from a Year 9 Maths textbook )

18. ## Re: HSC 2017 Maths (Advanced) Marathon

Show that x^3 - 8 has only one linear factor over the real number field,

What is the real number field?

19. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by 19KANguy
Show that x^3 - 8 has only one linear factor over the real number field,

What is the real number field?
To solve it in a different way, we can look at the roots, since $x^{3}-8$ is a cubic polynomial then there are three roots, $\alpha, \beta, \gamma$.Obviously one of the roots is 2,so let $\alpha = 2$
then sum and product of the roots imply
$\beta+ \gamma = -2, \beta \gamma = 4$
then
$( \beta+ \gamma)^{2} = \beta \gamma = 4$
which implies
$\beta^{2}+\gamma^{2} = -4$
and no real number satisfies above equality, hence there is only one linear factor

20. ## Re: HSC 2017 Maths (Advanced) Marathon

Hey guys,
I've got little to no idea about how to solve the logarithmic equation questions I've come across in the past hsc papers.

Can someone help me solve: 2lnx=ln(5+4x) AND 2log(base5)^3=log(base5)^x-log(base5)^6

21. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by dragon658
Hey guys,
I've got little to no idea about how to solve the logarithmic equation questions I've come across in the past hsc papers.

Can someone help me solve: 2lnx=ln(5+4x) AND 2log(base5)^3=log(base5)^x-log(base5)^6
\noindent \textbf{QUESTION 1} \\ \begin{align*} 2\ln{x} &= \ln{(5+4x)} \\ \ln{x^2} &= \ln{(5+4x)} \\ x^2 &= 5 + 4x \\ x^2 - 4x - 5 &= 0 \\ (x-5)(x+1) &= 0 \\ x &= -1, 5 \\ x \neq -1 \textrm{ } (x > 0), \textrm{ } x &= 5 \end{align*} \\\\\\ \textbf{QUESTION 2} \\ \begin{align*} 2\log_5{3} &= \log_5{x} - \log_5{6} \\ \log_5{9} &= \log_5{\frac{x}{6}} \\ 9 &= \frac{x}{6} \\ x &= 54 \end{align*}

22. ## Re: HSC 2017 Maths (Advanced) Marathon

Herooo, Pls help with this
Capture1.PNG

23. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by Mathew587
Herooo, Pls help with this
Capture1.PNG
$\noindent T_n = S_n - S_{n-1}. Your answer is T_{16} = 249 - 233 = 16$

24. ## Re: HSC 2017 Maths (Advanced) Marathon

\begin{align*}T_{50}&=S_{50}-S_{49}\\&= 249-233\\&= 16\end{align*}

25. ## Re: HSC 2017 Maths (Advanced) Marathon

Oh you can do that? My teachers a bit weird and would want us to actually write the whole solution out. How could you prove Tn = Sn - Sn-1?? ^^D

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