ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: tram on October 09, 2010, 04:02:50 pm

Title: Sequences/series questions
Post by: tram on October 09, 2010, 04:02:50 pm: hey guys,

my teacher gave me this whole compliation of spech multiple choice questions and had some real trouble with two sequences/series questions and wondering if anyone could help:

1) 4/3 - 1 + 3/4 - ........... is equal to

   a) 13/12
   b) 25/8
   c) 16/3
   d) 16/21
   e) 1

AND

2) the sum of the first 50 multiples of 7 is:

   a) 350
   b) 8925
   c) 9350
   d) 11025
   e) 7.88 x 10^15

And help would be greatly appreciated :D
Title: Re: Sequences/series questions
Post by: iNerd on October 09, 2010, 04:08:22 pm: If the answer to Q2 is (B); 8925 I will put up my workings
Title: Re: Sequences/series questions
Post by: Martoman on October 09, 2010, 04:22:22 pm: so the first is a sum to infinity.

You know $a = \frac{4}{3}$ and the common ratio is $r = \frac{-3}{4}$ . a is the first term and r is the common multiple you multiply the previous term by to get the next one.

The sum to infinity is $\frac{a}{1-r}$

So $\frac{\frac{4}{3}}{1-\frac{-3}{4}}} = \frac{\frac{4}{3}}{1+\frac{3}{4}}} =\frac{\frac{4}{3}}{\frac{7}{4}}}$

$\frac{4}{3} \times \frac{4}{7} = \frac{16}{21}$

2)

$7+14+21... 350$
$7(1+2+3+4...50)$

The $1+2+3+4...50$ forms an arithmetic series the formula is $\frac{n}{2}(a+l)$ where $n$ is the number of terms, $a$ is the first term and $l$ is the last term.

In general for n number of terms between $[a,b]$ there is $b-a +1$ number of terms. Think about it in the small scale ie how many numbers between 0 and 9? its 10, that is $9-0+1 = 10$ another example is how many numbers from 2 to 7 = $7-2+1 = 6.$ This *may* go against your intuition of just 7-2 which is why I am explaining this in detail so you don't get confused. Its because subtraction is a span and doesn't account for the end number.

So to finish this off $n = 50-1+1 = 50, a = 1, l = 50$

$\frac{50}{2}(1+50)$

$25(51) = 1275$

Finally $7(1+2+3...50) = 7(1275) = 8925$
Title: Re: Sequences/series questions
Post by: iNerd on October 09, 2010, 04:27:13 pm: Okay seems like my answer of 8925 was right and yet I did it differently to the legendary Martoman but then again I'm in Year 10 so I have no idea which way is better :P

First we identify that it is an arithmetic sequence and then we draw upon the formula: S_n = n/2[2a+(n-1)d] where:
a stands for the first term
d stands for the common ~~denominator~~ difference
n stands for the number of terms

In the question a = 7, d = 7 and n = 50
Sub these values into the formula and you will get 8925; B

EDIT: Martoman pointed out a subtle mistake :P
Title: Re: Sequences/series questions
Post by: stonecold on October 09, 2010, 04:27:31 pm: ^That is further maths lol... not that I can talk. I'd still manage to get it wrong. :P
Title: Re: Sequences/series questions
Post by: tram on October 09, 2010, 04:59:02 pm: cheers guys, relly appreciate the help (esp you martoman) NOW is see what question one was on about.... i couldn't even recognise the geometric patern lol >.<
Title: Re: Sequences/series questions
Post by: tram on October 09, 2010, 05:02:36 pm: Quote from: Martoman on October 09, 2010, 04:22:22 pm
so the first is a sum to infinity.

You know $a = \frac{4}{3}$ and the common ratio is $r = \frac{-3}{4}$ . a is the first term and r is the common multiple you multiply the previous term by to get the next one.

The sum to infinity is $\frac{a}{1-r}$

So $\frac{\frac{4}{3}}{1-\frac{-3}{4}}} = \frac{\frac{4}{3}}{1+\frac{3}{4}}} =\frac{\frac{4}{3}}{\frac{7}{4}}}$

$\frac{4}{3} \times \frac{4}{7} = \frac{16}{21}$

2)

$7+14+21... 350$
$7(1+2+3+4...50)$

The $1+2+3+4...50$ forms an arithmetic series the formula is $\frac{n}{2}(a+l)$ where $n$ is the number of terms, $a$ is the first term and $l$ is the last term.

In general for n number of terms between $[a,b]$ there is $b-a +1$ number of terms. Think about it in the small scale ie how many numbers between 0 and 9? its 10, that is $9-0+1 = 10$ another example is how many numbers from 2 to 7 = $7-2+1 = 6.$ This *may* go against your intuition of just 7-2 which is why I am explaining this in detail so you don't get confused. Its because subtraction is a span and doesn't account for the end number.

So to finish this off $n = 50-1+1 = 50, a = 1, l = 50$

$\frac{50}{2}(1+50)$

$25(51) = 1275$

Finally $7(1+2+3...50) = 7(1275) = 8925$

wow that is the first time sequences has made sense to me.....thx!

Quote from: iNerd on October 09, 2010, 04:27:13 pm
Okay seems like my answer of 8925 was right and yet I did it differently to the legendary Martoman but then again I'm in Year 10 so I have no idea which way is better :P

First we identify that it is an arithmetic sequence and then we draw upon the formula: S_n = n/2[2a+(n-1)d] where:
a stands for the first term
d stands for the common denominator
n stands for the number of terms

In the question a = 7, d = 7 and n = 50
Sub these values into the formula and you will get 8925; B

lol that's probs what i would've done, thx!
Title: Re: Sequences/series questions
Post by: Martoman on October 09, 2010, 10:23:41 pm: Quote from: iNerd on October 09, 2010, 04:27:13 pm
Okay seems like my answer of 8925 was right and yet I did it differently to the legendary Martoman but then again I'm in Year 10 so I have no idea which way is better :P

First we identify that it is an arithmetic sequence and then we draw upon the formula: S_n = n/2[2a+(n-1)d] where:
a stands for the first term
d stands for the common denominator
n stands for the number of terms

In the question a = 7, d = 7 and n = 50
Sub these values into the formula and you will get 8925; B

The $2a+(n-1)d = a + (a +(n-1)d)$

That thing in brackets, the $(a+(n-1)d)$ is the formula for an arithmetic sequence, it is used to find the last term. My method just simplifies it to (a+l) as we have both the first and last terms.

Quote from: iNerd on October 09, 2010, 04:27:13 pm
Okay seems like my answer of 8925 was right and yet I did it differently to the legendary Martoman but then again I'm in Year 10 so I have no idea which way is better :P

First we identify that it is an arithmetic sequence and then we draw upon the formula: S_n = n/2[2a+(n-1)d] where:
a stands for the first term
d stands for the common denominator
n stands for the number of terms

In the question a = 7, d = 7 and n = 50
Sub these values into the formula and you will get 8925; B

and tut tut, i should be in year 9 and I know that its a difference not a denominator :smitten: :smitten:
Title: Re: Sequences/series questions
Post by: Linkage1992 on October 10, 2010, 12:50:29 am: Quote
i should be in year 9

what ?! how old are you martoman??