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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: chikka on October 11, 2012, 06:23:36 pm

Title: Kilbaha 2012 Number Patterns Q
Post by: chikka on October 11, 2012, 06:23:36 pm

Question 8 on the multi choice

n-1, 4n-4 and 12n+16 are the first three terms of a geometric sequence.
The sum of the first four terms is:

A. 595
B. 634
C. 672
D. 728
E. 734

I know there must be some simple solution to this or some weird method for working it out, but I'm completely stuck :(
Any help will be muchly appreciated!!

Title: Re: Kilbaha 2012 Number Patterns Q
Post by: FlorianK on October 11, 2012, 06:41:58 pm

Quote from: chikka on October 11, 2012, 06:23:36 pm

n-1, 4n-4 and 12n+16 are the first three terms of a geometric sequence.
The sum of the first four terms is:

A. 595
B. 634
C. 672
D. 728
E. 734

(4n-4)/(n-1)=(12n+16)/(4n-4)
-> n= 8
-->7,28,112,448
-->595 -->A

Is that correct?

Title: Re: Kilbaha 2012 Number Patterns Q
Post by: chikka on October 11, 2012, 06:47:32 pm

Unfortunately don't have the solutions yet, but should be getting them tomorrow :)

Title: Re: Kilbaha 2012 Number Patterns Q
Post by: FlorianK on October 11, 2012, 06:49:51 pm

Assuming I'm correct:
Do you know how I got that answer?

Title: Re: Kilbaha 2012 Number Patterns Q
Post by: Special At Specialist on October 11, 2012, 07:08:37 pm

Quote from: FlorianK on October 11, 2012, 06:49:51 pm

Assuming I'm correct:
Do you know how I got that answer?

Yes. You set up a ratio problem by claiming that the ratio of the 1st term to the 2nd term will be the same as the ratio of the 2nd term to the 3rd term. You then solved the ratio to get the value n = 8 and then substituted that in to find the first 3 terms. After finding the first 3 terms, you used ratios again to tell that the 4th term is equal to 4 times the 3rd term (because that is what the ratio is between consecutive terms) and then added them all up to get 595 as your answer.