Kilbaha 2012 Number Patterns Q

[chikka]

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!!

[FlorianK]

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?

[chikka]

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

[FlorianK]

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

[Special At Specialist]

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.