If k+1 2k-1 and 3k+1 are three consecutive term of a Gp find the value of the common ratio
To find the common ratio (r) of a geometric progression (GP) given three consecutive terms, we can set up the following equation:
k + 1, 2k - 1, 3k + 1
The first term is k + 1, the second term is 2k - 1, and the third term is 3k + 1.
For a GP, we know that the ratio between consecutive terms is the same. Therefore, we can set up the following equation:
(2k - 1) / (k + 1) = (3k + 1) / (2k - 1)
To solve this equation, we can cross-multiply:
(2k - 1) * (2k - 1) = (k + 1) * (3k + 1)
Expanding both sides of the equation:
4k² - 4k + 1 = 3k² + k + 3k + 1
Combining like terms:
4k² - 4k + 1 = 3k² + 4k + 1
Rearranging the equation:
4k² - 3k² - 4k + 4k + 1 - 1 = 0
Simplifying:
k² = 0
Taking the square root of both sides:
k = 0
Therefore, the value of the common ratio (r) is:
r = (2k - 1) / (k + 1)
r = (2 * 0 - 1) / (0 + 1)
r = -1
So, the common ratio (r) of the geometric progression is -1.