Author Topic: Mathematics Question Thread (Read 1305868 times) Tweet Share

teapancakes08 · « **Reply #900 on:** November 12, 2016, 09:06:36 am »

Need to check for this question:

25. The sum of the first six terms of an arithmetic sequence is -12 and the sum of it's first fourteen term is 196. Find (a) in the sum of n terms, (b) the smallest value of n if the sum is to exceed 250.

I got the part a fine by using the formula Sn = (n/2)[2a + (n-1)d] and simultaneous equations, but the answer I got was 2n^2 - 12n....and the answer at the back is 2n^2 - 14n.

In part b I'm having trouble following. I tried using quadradics but with either one I end up getting surds (which is probably the point since 250 isn't a factor that fits nicely with either of them). And I don't think it possible to do logs because there are three terms...

If anyone could help that would be much appreciated, thanks ^^

RuiAce · « **Reply #901 on:** November 12, 2016, 09:21:48 am »

Quote from: teapancakes08 on November 12, 2016, 09:06:36 am

Need to check for this question:

25. The sum of the first six terms of an arithmetic sequence is -12 and the sum of it's first fourteen term is 196. Find (a) in the sum of n terms, (b) the smallest value of n if the sum is to exceed 250.

I got the part a fine by using the formula Sn = (n/2)[2a + (n-1)d] and simultaneous equations, but the answer I got was 2n^2 - 12n....and the answer at the back is 2n^2 - 14n.

In part b I'm having trouble following. I tried using quadradics but with either one I end up getting surds (which is probably the point since 250 isn't a factor that fits nicely with either of them). And I don't think it possible to do logs because there are three terms...

If anyone could help that would be much appreciated, thanks ^^

$\text{With correct calculations, the following system of equations:}\\ \begin{align*}S_6=\frac62(2a+5d)&=-12\\ S_{14}=\frac{14}{2}(2a+13d)&=196\end{align*}\\ \text{yields }a=-12, d=4$
$\text{To get an expression for a general }S_n\text{, upon substituting back in:}\\ \begin{align*}S_n&=\frac{n}{2}(-24+(n-1)4)\\ &= \frac{n}{2}(4n-28)\\ &=2n^2-14\end{align*}\\ \textbf{Check that you didn't make an algebra mistake somewhere}$
$\text{In b), note that what we }\textbf{actually}\text{ want to solve is }\\ S_n > 250\\ \text{We just replace > with = because it looks neater that way.}$
$\text{And we now consider the quadratic.}\\ \begin{align*}S_n&=250\\ 2n^2-14n-250&=0\\ n^2-7n-125&=0\\ n&=\frac{7\pm \sqrt{49+500}}{2}\\ &= -8.215..., 15.215...\end{align*}\\ \text{Indeed, the need for the quadratic formula is NOT surprising.}$
$\text{Clearly }n\text{ can't be a negative number, so we have }n=15.215...\\ \text{So because we want it to }\textbf{exceed}\text{, clearly we must go up by 1.}\\ \text{So the answer is }n=16$
$\textbf{Note that solving the inequality }S_n > 250\textbf{ makes it 'mathematically' tidier and more fluent}\\ \text{But it becomes visually less appealing, because it's a quadratic inequality nonetheless}$

jakesilove · « **Reply #902 on:** November 12, 2016, 09:23:12 am »

Quote from: teapancakes08 on November 12, 2016, 09:06:36 am

Need to check for this question:

25. The sum of the first six terms of an arithmetic sequence is -12 and the sum of it's first fourteen term is 196. Find (a) in the sum of n terms, (b) the smallest value of n if the sum is to exceed 250.

I got the part a fine by using the formula Sn = (n/2)[2a + (n-1)d] and simultaneous equations, but the answer I got was 2n^2 - 12n....and the answer at the back is 2n^2 - 14n.

In part b I'm having trouble following. I tried using quadradics but with either one I end up getting surds (which is probably the point since 250 isn't a factor that fits nicely with either of them). And I don't think it possible to do logs because there are three terms...

If anyone could help that would be much appreciated, thanks ^^

Hey! Just setting n=6, you'll see that the only equation that gets out the required value of -12 is the one at the back of the book. Potentially you messed up some algebra somewhere? Check it over, and if you still need help, post it here and I'll take a look!

As for the second part, you're 100% on the right track. Set your equation equal to 250, and find the surd value. Using the equation at the back of the book, n will equal -8.215 and 15.215. Now, obviously n can't be zero, so we take the positive value 15.215. Let's think about what that means for a second; if n=15, the answer will be LESS than 250. If n=16, the answer will be MORE than 250. So, 16 is out answer, because that's what the question wants!

Let me know if that all makes sense.

Jake

teapancakes08 · « **Reply #903 on:** November 12, 2016, 09:45:28 am »

Quote from: jakesilove on November 12, 2016, 09:23:12 am

Hey! Just setting n=6, you'll see that the only equation that gets out the required value of -12 is the one at the back of the book. Potentially you messed up some algebra somewhere? Check it over, and if you still need help, post it here and I'll take a look!

As for the second part, you're 100% on the right track. Set your equation equal to 250, and find the surd value. Using the equation at the back of the book, n will equal -8.215 and 15.215. Now, obviously n can't be zero, so we take the positive value 15.215. Let's think about what that means for a second; if n=15, the answer will be LESS than 250. If n=16, the answer will be MORE than 250. So, 16 is out answer, because that's what the question wants!

Let me know if that all makes sense.

Jake

I think I got it. I checked back and saw that I made an algebra mistake when subbing in the values for a and d. Ruiace also helped me out on that one. As for the part b, I should be good. The explanations helped clear things up a lot. Thanks so much for the help ^^

teapancakes08 · « **Reply #904 on:** November 12, 2016, 09:59:38 am »

Quote from: RuiAce on November 12, 2016, 09:21:48 am

$\text{With correct calculations, the following system of equations:}\\ \begin{align*}S_6=\frac62(2a+5d)&=-12\\ S_{14}=\frac{14}{2}(2a+13d)&=196\end{align*}\\ \text{yields }a=-12, d=4$
$\text{To get an expression for a general }S_n\text{, upon substituting back in:}\\ \begin{align*}S_n&=\frac{n}{2}(-24+(n-1)4)\\ &= \frac{n}{2}(4n-28)\\ &=2n^2-14\end{align*}\\ \textbf{Check that you didn't make an algebra mistake somewhere}$
$\text{In b), note that what we }\textbf{actually}\text{ want to solve is }\\ S_n > 250\\ \text{We just replace > with = because it looks neater that way.}$
$\text{And we now consider the quadratic.}\\ \begin{align*}S_n&=250\\ 2n^2-14n-250&=0\\ n^2-7n-125&=0\\ n&=\frac{7\pm \sqrt{49+500}}{2}\\ &= -8.215..., 15.215...\end{align*}\\ \text{Indeed, the need for the quadratic formula is NOT surprising.}$
$\text{Clearly }n\text{ can't be a negative number, so we have }n=15.215...\\ \text{So because we want it to }\textbf{exceed}\text{, clearly we must go up by 1.}\\ \text{So the answer is }n=16$
$\textbf{Note that solving the inequality }S_n > 250\textbf{ makes it 'mathematically' tidier and more fluent}\\ \text{But it becomes visually less appealing, because it's a quadratic inequality nonetheless}$

Guess I did make an algebraic error ^^; but I suppose better now than during the test.

Thanks for the explanation, it cleared my understanding of it by a large margin. Thank you so much for the help ^^

julzzz · « **Reply #905 on:** November 12, 2016, 10:45:12 am »

(1/27)^(3n+1)=sqrt{3}/81

This is a prelim question which i couldnt manage to do
Using indices, not log, thanks

jakesilove · « **Reply #906 on:** November 12, 2016, 11:01:55 am »

Quote from: julzzz on November 12, 2016, 10:45:12 am

(1/27)^(3n+1)=sqrt{3}/81

This is a prelim question which i couldnt manage to do
Using indices, not log, thanks

So, you're question is

$(\frac{1}{27})^{3n+1}=\frac{\sqrt{3}}{81}$

We need to convert everything into powers of three;

$(\frac{1}{3^3})^{3n+1}=\frac{3^{0.5}}{3^4}$

$(3^{-3})^{3n+1}=3^{0.5}*3^{-4}$

$3^{-9n-3}=3^{-3.5}$

Therefore, we can let

$-9n-3=-3.5$

$n=\frac{0.5}{9}$

So our answer is n=1/18! Hope that all makes sense

teapancakes08 · « **Reply #907 on:** November 12, 2016, 09:51:00 pm »

Wondering about this question:

20. Show that the sum of the first n odd natural numbers is a perfect square.

The method I've tried is similar to that of induction, letting the sum equal to n^2 and solving it by plugging in terms, where n = the number of terms in current series and the difference being two...if that makes any sense (as in 1 = 1^2, 1 + 3 = 2^2....etc.). Is there a way to neatly show this without confusing yourself (and, by extension, probably the marker)? Or is there a better way to express this proof in general?

RuiAce · « **Reply #908 on:** November 12, 2016, 09:54:22 pm »

Quote from: teapancakes08 on November 12, 2016, 09:51:00 pm

Wondering about this question:

20. Show that the sum of the first n odd natural numbers is a perfect square.

The method I've tried is similar to that of induction, letting the sum equal to n^2 and solving it by plugging in terms, where n = the number of terms in current series and the difference being two...if that makes any sense (as in 1 = 1^2, 1 + 3 = 2^2....etc.). Is there a way to neatly show this without confusing yourself (and, by extension, probably the marker)? Or is there a better way to express this proof in general?

$\text{Just consider }S_n=1+3+5+\dots+(2n-1)\\ \text{The idea is to consider an A.P. with }a=1, \,d=2$
$\text{So by summing the A.P. using }S_n=\frac{n}{2}(a+\ell)\\ \begin{align*}S_n&=\frac{n}{2}(1+(2n-1))\\ &=n^2\end{align*}$

RuiAce · « **Reply #909 on:** November 12, 2016, 09:57:53 pm »

samuels1999 · « **Reply #910 on:** November 12, 2016, 11:15:54 pm »

Hi everyone

I got this question. And it is nothing like the usual ones I have done before. It is probably really easy, but I found it a little tricky.

Thanks,
Samuel

RuiAce · « **Reply #911 on:** November 13, 2016, 06:03:59 am »

Quote from: samuels1999 on November 12, 2016, 11:15:54 pm

Hi everyone

I got this question. And it is nothing like the usual ones I have done before. It is probably really easy, but I found it a little tricky.

Thanks,
Samuel

$\text{For a), if the score happened to be 1}\\ \text{Analysing the scenario, this can ONLY occur}\\ \text{if both dice ended up with a 1.}$
$\text{The probability of (1,1) is clearly }\frac{1}{36}$
________________________
$\text{Essentially if we want three consecutive scores of 1}\\ \text{Not only do we have to get }(1,1)\text{ in the first throw}\\ \text{But we must get it in the subsequent throws as well.}\\ \text{So we have }\frac{1}{36}\times\frac{1}{36}\times\frac{1}{36}$
________________________
$\text{Analyse the scenario carefully.}\\ \text{For c), if the score were to be a 6, then:}$
$\textit{Whilst both dice CAN show a 6}\\ \textit{We only care that at least ONE of the dice shows a 6}$
$\text{There are many ways of tackling this.}\\ \text{The easiest way is to consider the }\textit{complement}\\ \text{(This is due to the presence of 'at least')}$
$\begin{align*}Pr(\text{At least 1 six})&=1-Pr(\text{No sixes})\\ &= 1-\frac{25}{36}\\ &= \frac{11}{36}\end{align*}$
$\text{Note: }\frac{25}{36}\text{ comes from the fact that}\\ \text{Probability of an event}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}\\ \text{Think about why favourable outcomes = 25}$

Aussie1Italia2 · « **Reply #912 on:** November 13, 2016, 09:42:21 am »

Hello, please and thank you for your help!

Alpha & beta are roots of the quadratic equation x² - 6x + 3 = 0.

Find alpha x beta² + alpha² x beta

jamonwindeyer · « **Reply #913 on:** November 13, 2016, 09:59:30 am »

Quote from: Aussie1Italia2 on November 13, 2016, 09:42:21 am

Hello, please and thank you for your help!

Alpha & beta are roots of the quadratic equation x² - 6x + 3 = 0.

Find alpha x beta² + alpha² x beta

Hey Aussie!

$\alpha+\beta=-\frac{b}{a}=6\\\alpha\beta=\frac{c}{a}=3\\\alpha\beta^2+\alpha^2\beta=\alpha\beta(\alpha+\beta)$

And then you just sub and go to get 18 as your answer

katnisschung · « **Reply #914 on:** November 13, 2016, 11:55:58 am »

I probably got it completely wrong but don't know becos they're arent any answers

Q) ap: 52,46,40
Find the smallest value of n so Sn<0
so i got n=19

i couldn't get it any bigger becos the limit for file size

Moderator Edit: Merged question w/ answer

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