If k+1,2k-1and 3k+1 are three consecutive terms of a GP ,find the value of the common ratio
In a geometric progression (GP), each term is found by multiplying the previous term by a constant factor called the common ratio.
So, let's find the common ratio, r, by using the given terms:
k + 1, 2k - 1, 3k + 1
To check if these terms are in a GP, we need to see if the ratio of any two consecutive terms is the same.
First, let's find the ratio of the second term to the first term:
(2k - 1)/(k + 1)
Next, let's find the ratio of the third term to the second term:
(3k + 1)/(2k - 1)
Since the three terms are consecutive terms in a GP, both ratios above should be equal to the common ratio, r.
Setting up the equation:
(2k - 1)/(k + 1) = (3k + 1)/(2k - 1)
Cross-multiplying:
(2k - 1)(2k - 1) = (k + 1)(3k + 1)
Expanding both sides:
4k^2 - 4k + 1 = 3k^2 + 4k + 1
Simplifying:
4k^2 - 4k + 1 - 3k^2 - 4k - 1 = 0
Combining like terms:
k^2 - 8k = 0
Factoring out k:
k(k - 8) = 0
Setting each factor equal to zero and solving for k:
k = 0 or k - 8 = 0
If k = 0, then the terms would be 1, -1, and 1, which is not a GP.
If k - 8 = 0, then k = 8.
Therefore, the value of the common ratio, r, is 8.