HSC 2016 Maths Marathon (archive) (4 Viewers)

Paradoxica · Jun 19, 2016

Re: HSC 2016 2U Marathon

Leehuan is correct, this question contains insufficient information to deduce the value of r analytically.

Unless you're expected to guess... I guess....

davidgoes4wce · Jun 19, 2016

Re: HSC 2016 2U Marathon

This is the page from the book if you guys are keen:

My thinking is they have a little picture of a calculator next to it, so we have to use some sort of trial and error to solve for 'r'.

si2136 · Jun 19, 2016

Re: HSC 2016 2U Marathon

davidgoes4wce said:
This is the page from the book if you guys are keen:

My thinking is they have a little picture of a calculator next to it, so we have to use some sort of trial and error to solve for 'r'.

Yep

InteGrand · Jun 25, 2016

Re: HSC 2016 2U Marathon

NEW QUESTION

$\noindent Show that the average of the numbers in a finite list of (at least two) numbers in arithmetic progression is equal to the average of the smallest and largest terms.$

Nailgun · Jun 25, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
NEW QUESTION

$\noindent Show that the average of the numbers in a finite list of (at least two) numbers in arithmetic progression is equal to the average of the smallest and largest terms.$

$$\noindent For the sum of an arbitrary list of $n$ numbers in a finite arithmetic sequence we may use the expression ${ S }_{ n }=\frac { n }{ 2 } (a+l)$, where the smallest number is $a$ and the largest is $l$.$

$$\noindent The average of the smallest and largest numbers will be$ \frac { a+l }{ 2 }$

$\noindent The average of any arbitrary finite list will be $\frac { { S }_{ n } }{ n }$

$\noindent Hence we are to show $\frac { \frac { n }{ 2 } (a+l) }{ n } =\frac { a+l }{ 2 }$

$$\noindent $LHS$: $\frac { \frac { n }{ 2 } (a+l) }{ n } =\frac { a+l }{ 2 } $. Therefore $LHS$=$RHS$.$

InteGrand · Jun 25, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
$$\noindent For the sum of an arbitrary list of $n$ numbers in a finite arithmetic sequence we may use the expression ${ S }_{ n }=\frac { n }{ 2 } (a+l)$, where the smallest number is $a$ and the largest is $l$.$

$$\noindent The average of the smallest and largest numbers will be$ \frac { a+l }{ 2 }$

$\noindent The average of any arbitrary finite list will be $\frac { { S }_{ n } }{ n }$

$\noindent Hence we are to show $\frac { \frac { n }{ 2 } (a+l) }{ n } =\frac { a+l }{ 2 }$

$$\noindent $LHS$: $\frac { \frac { n }{ 2 } (a+l) }{ n } =\frac { a+l }{ 2 } $. Therefore $LHS$=$RHS$.$

Well done!

davidgoes4wce · Jun 26, 2016

Re: HSC 2016 2U Marathon

Assuming an arithmetic series,

are HSC students expected to know something along the lines of say

$T_{n+1}=T_n+4$

to say something like

$T_{n-1}=T_{n-2}+4$

?

InteGrand · Jun 26, 2016

Re: HSC 2016 2U Marathon

davidgoes4wce said:
Assuming an arithmetic series,

are HSC students expected to know something along the lines of say

$T_{n+1}=T_n+4$

to say something like

$T_{n-1}=T_{n-2}+4$

?

Well it's basically just the same thing.

davidgoes4wce · Jun 26, 2016

Re: HSC 2016 2U Marathon

Yeah I want to check whether the students are expected to know it?

Nailgun · Jun 26, 2016

Re: HSC 2016 2U Marathon

davidgoes4wce said:
Assuming an arithmetic series,

are HSC students expected to know something along the lines of say

$T_{n+1}=T_n+4$

to say something like

$T_{n-1}=T_{n-2}+4$

?

I would say so

dan964 · Jun 27, 2016

Re: HSC 2016 2U Marathon

davidgoes4wce said:
Assuming an arithmetic series,

are HSC students expected to know something along the lines of say

$T_{n+1}=T_n+4$

to say something like

$T_{n-1}=T_{n-2}+4$

?

Yes, I guess.
If $T_{n}$ is defined as such for I presume the natural numbers, then $T_{n-2}$ has to be defined for n>2.

leehuan · Jun 27, 2016

Re: HSC 2016 2U Marathon

davidgoes4wce said:
Assuming an arithmetic series,

are HSC students expected to know something along the lines of say

$T_{n+1}=T_n+4$

to say something like

$T_{n-1}=T_{n-2}+4$

?

Without breaking restrictions on n the fact that you can move the term number is elementary assumed knowledge for 2U HSC

InteGrand · Jul 4, 2016

Re: HSC 2016 2U Marathon

NEW QUESTION

$\noindent Suppose that a sequence of real numbers is both an A.P. and a G.P. Show that the sequence is a constant sequence.$

Nailgun · Jul 4, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
NEW QUESTION

$\noindent Suppose that a sequence of real numbers is both an A.P. and a G.P. Show that the sequence is a constant sequence.$

$\noindent Let any ordered set of 3 consecutive terms in the sequence be $a,b,c$

$b-a=c-b\\ b=\frac { a+c }{ 2 } \\ \\ \frac { b }{ a } =\frac { c }{ b } \\ \\ { b }^{ 2 }=ac\\ \\ \\ \therefore \frac { { (a+c) }^{ 2 } }{ 4 } =ac\\ \\ { (a-c) }^{ 2 }=0\\ a=c\\ \\ b=\frac { c+c }{ 2 } =c\\ \\ \therefore a=b=c\\ \\ $Therefore, it is a constant sequence$$

leehuan · Jul 4, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
NEW QUESTION

$\noindent Suppose that a sequence of real numbers is both an A.P. and a G.P. Show that the sequence is a constant sequence.$

Just out of curiosity, could you potentially argue this from the equality of AM-GM being when all terms are the same?

Paradoxica · Jul 4, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Just out of curiosity, could you potentially argue this from the equality of AM-GM being when all terms are the same?

Yes.

leehuan · Jul 9, 2016

Re: HSC 2016 2U Marathon

Simplify (x-a)(x-b)...(x-z)

InteGrand · Jul 9, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Simplify (x-a)(x-b)...(x-z)

Lol, I remember this was on Terry Lee's website. Funny Q.

jathu123 · Jul 9, 2016

Re: HSC 2016 2U Marathon

haha ye remember it from this youtube vid, funny indeed lol

Green Yoda · Jul 9, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Simplify (x-a)(x-b)...(x-z)

Not sure if this is correct but if its the whole alphabet then (x-x)=0 and since its a factor the whole expression = 0?

HSC 2016 Maths Marathon (archive) (4 Viewers)

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

Cole World

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Cole World

what

Well-Known Member

Well-Known Member

Cole World

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Active Member

Hi Φ

Users Who Are Viewing This Thread (Users: 0, Guests: 4)