Author Topic: Mathematics Question Thread (Read 1626444 times) Tweet Share

teapancakes08 · « **Reply #930 on:** November 15, 2016, 07:45:46 pm »

16. The population of a certain town is 10000; if its population will decrease each year by 10% of it population in the preceding year, find its population in 5 years time.

Which GS formula would I use to solve this one? I tried using both and got the really weird answers...to which I take it that I'm looking at the question wrong...

Any suggestions on how to approach it?

jakesilove · « **Reply #931 on:** November 15, 2016, 09:09:58 pm »

Quote from: teapancakes08 on November 15, 2016, 07:45:46 pm

16. The population of a certain town is 10000; if its population will decrease each year by 10% of it population in the preceding year, find its population in 5 years time.

Which GS formula would I use to solve this one? I tried using both and got the really weird answers...to which I take it that I'm looking at the question wrong...

Any suggestions on how to approach it?

To be honest, I'd probably do this without bothering with the series, because doing the iteration 5 times isn't too difficult.

The first year, there are 10,000 people. Then next year (after one year), there will be 0,9*10,000=9,000 people. The next year (after two years), there will be 0,9*9,000=8,100 people. The next year (after three years) there will be 0.9*8,100=7,290 people. The next year (after four years), there will be 0.9*7290=6561 people. The next year (after five years), there will be 0.9*6561=5904.7 people (which is either 5905 or 5904 people, depending on how you choose to round).

You could have set up a GP, which the common ratio being 0.9 and a=10,000. Works just as well. Difficulty would be whether you set n=5 or n=6; 6 gets you the answer I posited above

teapancakes08 · « **Reply #932 on:** November 15, 2016, 09:46:33 pm »

Quote from: jakesilove on November 15, 2016, 09:09:58 pm

To be honest, I'd probably do this without bothering with the series, because doing the iteration 5 times isn't too difficult.

The first year, there are 10,000 people. Then next year (after one year), there will be 0,9*10,000=9,000 people. The next year (after two years), there will be 0,9*9,000=8,100 people. The next year (after three years) there will be 0.9*8,100=7,290 people. The next year (after four years), there will be 0.9*7290=6561 people. The next year (after five years), there will be 0.9*6561=5904.7 people (which is either 5905 or 5904 people, depending on how you choose to round).

You could have set up a GP, which the common ratio being 0.9 and a=10,000. Works just as well. Difficulty would be whether you set n=5 or n=6; 6 gets you the answer I posited above

Thanks for clearing things up ^^

I went back and found out that using Tn = ar^n-1 worked for this equation. I think n would be 6 regardless as it's five years plus the first year (i found it after plugging in n = 5 and got the wrong answer). When plugging it into the formula it give 5904.5, but since you can't half a person I rounded it up to 5905. It gets the answer pretty fast too, although powering through five terms isn't too hard. Once again, thank you for the help

teapancakes08 · « **Reply #933 on:** November 15, 2016, 10:26:21 pm »

More GS questions...

23. Find the first three terms of a geometric sequence given that the sum of the first four terms is 21 2/3, and the sum to infinity is 27.

FallonXay · « **Reply #934 on:** November 15, 2016, 11:21:55 pm »

Quote from: teapancakes08 on November 15, 2016, 10:26:21 pm

More GS questions...

23. Find the first three terms of a geometric sequence given that the sum of the first four terms is 21 2/3, and the sum to infinity is 27.

So from the given information, you know that $\frac{a}{(1-r)}\ = 27$ (sum to infinity) and that

$\begin{equation} \frac{a(1-r^4)}{1-r}\ = \frac{65}{3}\ \end{equation} $ (sum of the first 4 terms of the GS)

Now, you want to eliminate one of the variables so that you can solve for the other variable. To do this you want to find a common subject, the easiest way in this case is to rearrange for the common subject of

$\begin{equation} \frac{a}{(1-r)}\ \end{equation}$

So rearranging the equation for the sum of the first 4 terms to this will give you

$ \begin{equation} \frac{a}{(1-r)}=\frac{\left(\dfrac{\overline{65}}{3}\right)}{1-r^4}\ \end{equation} $

Once you have achieved this common subject, you can equate the sum to infinity formula and the newly rearranged formula giving you:

$\frac{\left(\dfrac{\overline{65}}{3}\right)}{1-r^4}=27$

Then solving for r should give you a ratio of 2/3. Sub this back into one of your original equations (for example, the sum to infinity) and this should solve for 'a' giving you the value of 9. Then proceed to multiply this by your now known ratio of 2/3 to find the first 3 terms of the geometric series which will be 9, 6 and 4.

Hope this helps

(p.s trying out latex and I kind of eventually gave up so if something doesn't make sense feel free to ask )

RuiAce · « **Reply #935 on:** November 16, 2016, 08:17:41 am »

Quote from: FallonXay on November 15, 2016, 11:21:55 pm

(p.s trying out latex and I kind of eventually gave up so if something doesn't make sense feel free to ask )

If you wanna try LaTeX don't use my guide at the bottom of the math section for now. Try experimenting the basics with Daum Equation Editor and then when you're typing it if something seems weird ask Jamon or myself.

Anything's hard to learn but LaTeX is just better than word and has a ton of other advantages when you use it to write up pdfs that needs heaps of math equations. Can also be quite fun

FallonXay · « **Reply #936 on:** November 16, 2016, 09:00:01 am »

Quote from: RuiAce on November 16, 2016, 08:17:41 am

If you wanna try LaTeX don't use my guide at the bottom of the math section for now. Try experimenting the basics with Daum Equation Editor and then when you're typing it if something seems weird ask Jamon or myself.

Anything's hard to learn but LaTeX is just better than word and has a ton of other advantages when you use it to write up pdfs that needs heaps of math equations. Can also be quite fun

Ooooooh, I didn't even realise you had a guide!!! I'm gonna check it out now despite your word of warning hehehehe. Also, will check out Daum Equation Editor. Thanks for the advice

RuiAce · « **Reply #937 on:** November 16, 2016, 09:07:40 am »

Quote from: FallonXay on November 16, 2016, 09:00:01 am

Ooooooh, I didn't even realise you had a guide!!! I'm gonna check it out now despite your word of warning hehehehe. Also, will check out Daum Equation Editor. Thanks for the advice

Thing with it is that after learning some more LaTeX I realised I overcomplicated it. I'm gonna be fixing it but I probably won't get started on that until Dec

teapancakes08 · « **Reply #938 on:** November 16, 2016, 10:09:06 pm »

33. Find the sum of series 1 + 1/(a+1) + 1/(a+1)^2 +.... For what range of values of a has this infinite series a sum?

I found the sum, which (a+1)/a, but I don't quite understand what to do for the second part. Any suggestions?

RuiAce · « **Reply #939 on:** November 16, 2016, 10:13:51 pm »

Quote from: teapancakes08 on November 16, 2016, 10:09:06 pm

33. Find the sum of series 1 + 1/(a+1) + 1/(a+1)^2 +.... For what range of values of a has this infinite series a sum?

I found the sum, which (a+1)/a, but I don't quite understand what to do for the second part. Any suggestions?

$\text{For a geometric series, an infinite series exists}\\ \text{if and only if the common ratio }r\text{ satisfies }|r|< 1\\ \text{and }r\text{ is well defined (you don't need to properly know what this means)}$
$\text{Note that }r=\frac{1}{a+1}\\ \text{Hence from }|r|<1\\ \begin{align*}\left|\frac{1}{a+1}\right|&<1\\ \frac{1}{|a+1|}&< 1\\ |a+1|&> 1\\ a+1<-1 \,&\&\,a+1> 1\\ a<-2 \, &\&\, a > 0\end{align*}$
Blech. Cheers Jamon (y)

jamonwindeyer · « **Reply #940 on:** November 16, 2016, 10:15:45 pm »

Quote from: teapancakes08 on November 16, 2016, 10:09:06 pm

33. Find the sum of series 1 + 1/(a+1) + 1/(a+1)^2 +.... For what range of values of a has this infinite series a sum?

I found the sum, which (a+1)/a, but I don't quite understand what to do for the second part. Any suggestions?

Hey! So the idea here is that infinite series only converge when the absolute value of the common ratio is less than 1.

So, we form an absolute value inequality

$|\frac{1}{a+1}|<1\\\frac{1}{|a+1|}<1\\|a+1|>1\\\therefore a>0, a<-2$

Edit: Woops, double, but I'l leave it, Rui I think you made a sign error from Line 4 to Line 5

teapancakes08 · « **Reply #941 on:** November 16, 2016, 10:24:50 pm »

List of GS logs because I have little clue on how to do them...

34. For the geometric sequence a, ar, ar^2, ...show that the sequence loga, log (ar), log (ar^2), .... is an arithmetic sequence.

36. Find, without using tables or calculator, the sum of 7 terms of log27 + log9 + log3 + ....

RuiAce · « **Reply #942 on:** November 16, 2016, 10:33:09 pm »

Quote from: teapancakes08 on November 16, 2016, 10:24:50 pm

List of GS logs because I have little clue on how to do them...

34. For the geometric sequence a, ar, ar^2, ...show that the sequence loga, log (ar), log (ar^2), .... is an arithmetic sequence.

36. Find, without using tables or calculator, the sum of 7 terms of log27 + log9 + log3 + ....

$\text{By considering the following logarithm laws}\\ \begin{align*}\log(MN)&=\log M+\log N\\ \log(M^n) &= n \log M\end{align*}$
$\text{The sequence }\log a, \log (ar), \log (ar^2), \log (ar^3), \dots \\ \text{can be rewritten as}\\ \log a, \log a + \log r, \log a + 2\log r, \log a + 3\log r, \dots\\ \text{Which is clearly an arithmetic sequence with first term }\log a\text{ and common ratio }\log r$
_____________________________________________________
$\text{Observe that we can rewrite the sum as}\\ \begin{align*}\log 27 + \log 9 + \log 3 + \dots &= 3 \log 3 + 2 \log 3 + 1\log 3 + \dots\\ &=(3+2+1+\dots)\log 3\end{align*}\\ \text{and what's in the bracket is clearly an arithmetic series with }a=3, d=-1$
$\text{So the sum of the first 7 terms is}\\ S_7=\frac{7}{2}(2\log 3 + 6(-1))\\ \text{which, according to the question, you leave in simplified exact form }7(\log 3 - 3)$

teapancakes08 · « **Reply #943 on:** November 16, 2016, 11:07:04 pm »

One more log...

35. Find the sum of 10 terms of the series log3 + log6 + log12 + .... given that log 3 = 0.4771 and log2 = 0.3010

I'm sure I can figure it out (using logmn = logm + logn), but the answer I got was way off...and r was confusing to find...)

RuiAce · « **Reply #944 on:** November 16, 2016, 11:10:08 pm »

Quote from: teapancakes08 on November 16, 2016, 11:07:04 pm

One more log...

35. Find the sum of 10 terms of the series log3 + log6 + log12 + .... given that log 3 = 0.4771 and log2 = 0.3010

I'm sure I can figure it out (using logmn = logm + logn), but the answer I got was way off...and r was confusing to find...)

You can't treat these log ones like a geometric series. Because if you try putting everything inside one log you get this:
log(3*6*12*...)

Notice how it's all multiply. Not add. This is a product, not a sum.

Have to decompose it into
(log 3) + (log 3 + log 2) + (log 3 + 2 log 2) + ... which is an arithmetic series.

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