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, ar2, ar3, ..., arn, ...
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 tn/tn-1 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 nth term.
To determine two geometric sequences with the given first terms 18x-9, 2x+8, and x-1, we need to find the common ratio (r) and the base (a) of each sequence.
First, set up the equation (2x+8)/(18x-9) = (x-1)/(2x+8) to find the common ratio. Cross-multiply the equation to get:
(2x+8)*(2x+8) = (18x-9)*(x-1)
Expand and simplify the equation:
4x^2 + 32x + 64 = 18x^2 - 27x - 18
Combine like terms and move all terms to one side:
14x^2 - 59x - 82 = 0
Solve this quadratic equation to find the value(s) of x.
Once you have the value(s) of x, substitute them back into the first terms 18x-9, 2x+8, and x-1 to determine the actual values of the first terms.
After finding the first terms, you can use the common ratio to find the base. Recall that the common ratio (r) is found by taking the ratio of any term divided by its preceding term in the sequence.
Finally, with the base and the common ratio known, you can write the general or n-th term for each geometric sequence using the formula a, ar, ar^2, ar^3, ..., ar^n, ...
Remember to check your solution by plugging the values back into the original equation to ensure they satisfy both sides of the equation.