Question
determine two geometreic sequences whose first terms are 18x-9, 2x+8 and x-1
how would I go about and do this question?
I have no idea! so a little help would be really helpful! thanks!
A geometric sequence looks like
a, ar, ar<sup>2</sup>, ar<sup>3</sup>, ..., ar<sup>n</sup>, ...
where a is usually called the base and r is the common ratio.
If the first three terms are 18x-9, 2x+8 and x-1 then
(2x+8)/(18x-9) = (x-1)/(2x+8)
i.e., the ratio of t<sub>n</sub>/t<sub>n-1</sub> is the common ratio.
So solve
(2x+8)*(2x+8)=(18x-9)*(x-1) for x.
Then substitute the values of x into the first 3 terms to determine what they are, then follow the above to determine the common ratio, the base, then the general or n<sup>th</sup> term.
how would I go about and do this question?
I have no idea! so a little help would be really helpful! thanks!
A geometric sequence looks like
a, ar, ar<sup>2</sup>, ar<sup>3</sup>, ..., ar<sup>n</sup>, ...
where a is usually called the base and r is the common ratio.
If the first three terms are 18x-9, 2x+8 and x-1 then
(2x+8)/(18x-9) = (x-1)/(2x+8)
i.e., the ratio of t<sub>n</sub>/t<sub>n-1</sub> is the common ratio.
So solve
(2x+8)*(2x+8)=(18x-9)*(x-1) for x.
Then substitute the values of x into the first 3 terms to determine what they are, then follow the above to determine the common ratio, the base, then the general or n<sup>th</sup> term.