My Math Forum exam tomorrow, few problems with surds,sequences,factorising

 Algebra Pre-Algebra and Basic Algebra Math Forum

 January 9th, 2011, 03:01 AM #1 Newbie   Joined: Jan 2011 Posts: 11 Thanks: 0 exam tomorrow, few problems with surds,sequences,factorising Big day tomorrow! I have found some past paper questions, but there is no mark scheme and I don't get some of them! Firstly: 32timesroot 2=2 to power of a, find a
 January 9th, 2011, 03:03 AM #2 Newbie   Joined: Jan 2011 Posts: 11 Thanks: 0 Re: exam tomorrow, few problems with surds,sequences,factori -If some of you could please have a clear concise method of doing these it would be really appreciated, and not just answers!
 January 9th, 2011, 03:18 AM #3 Newbie   Joined: Jan 2011 Posts: 11 Thanks: 0 Re: exam tomorrow, few problems with surds,sequences,factori I'm thinking 32 to power of 1 times 2 to power of 1/2 =2 to power of a 1 1/2 =a?
 January 9th, 2011, 03:33 AM #4 Senior Member   Joined: Feb 2010 Posts: 135 Thanks: 0 Re: exam tomorrow, few problems with surds,sequences,factori $32sqrt2= 2^a$ $2^5sqrt2= 2^a$ 32 = $2^5$ $2^{5.5}$ = $2^a$ a = 5.5
 January 9th, 2011, 05:50 AM #5 Senior Member   Joined: Oct 2010 From: Vietnam Posts: 226 Thanks: 0 Re: exam tomorrow, few problems with surds,sequences,factori 32.squareroot2=2^a 2^5.2^(0.5)=2^a 2^(5.5)=2^a So a=5.5
January 9th, 2011, 07:58 AM   #6
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Re: exam tomorrow, few problems with surds,sequences,factori

Ok I understand now as 5+ 1/2 =5.5.

The next problem I have is this sequences problem as shown in the picture...

I factorized it given that a=7 and d=6, and ended up with 172 and that doesn't work...
Attached Images
 maths problem.jpg (24.9 KB, 382 views)

 January 9th, 2011, 08:12 AM #7 Newbie   Joined: Jan 2011 Posts: 22 Thanks: 0 Re: exam tomorrow, few problems with surds,sequences,factori Have a look here, it might help http://en.wikipedia.org/wiki/Arithmetic_progression
 January 9th, 2011, 08:22 AM #8 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: exam tomorrow, few problems with surds,sequences,factori Hello Gladiatorx, Chimoo's information is useful. Maybe it helps you understand: $\sum_{r=1}^n 6r+1=\sum_{r=1}^n 6r+\sum_{r=1}^n 1=6\sum_{r=1}^n r+\sum_{r=1}^n 1=6\cdot 0.5(n^2+n)+n=3n^2+4n$ Can you find n such that $3n^2+4n=2464$? Which solution for “n” do you choose and why? Hoempa
 January 9th, 2011, 08:34 AM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond Re: exam tomorrow, few problems with surds,sequences,factori The sum of an arithmetic series is given by $\frac{n}{2}\left[2a\,+\,(n\,-\,1)d\right]$ where $a$ is the first term and $d$ is the common difference. Substituting known values: $2464\,=\,\frac{n}{2}\left[2(7)\,+\,(n\,-\,1)6\right]$ $4928\,=\,n\left[14\,+\,6n\,-\,6\right]$ $4928\,=\,8n\,+\,6n^2$ $6n^2\,+\,8n\,-\,4928\,=\,0$ Solve and choose the appropriate root.
 January 9th, 2011, 10:17 AM #10 Newbie   Joined: Jan 2011 Posts: 11 Thanks: 0 Re: exam tomorrow, few problems with surds,sequences,factori i got to there and got 172, but using n=172 gives a massive answer so....

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