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, the ratio between consecutive terms is constant. Let's find the ratio between the second term and the first term, and the ratio between the third term and the second term.
The ratio between the second term (2k-1) and the first term (k+1) is (2k-1)/(k+1).
The ratio between the third term (3k+1) and the second term (2k-1) is (3k+1)/(2k-1).
Since the three terms are consecutive terms of a geometric progression, the two ratios should be equal:
(2k-1)/(k+1) = (3k+1)/(2k-1).
Cross-multiplying, we get:
(2k-1)(2k-1) = (k+1)(3k+1).
Expanding both sides:
4k^2 - 2(2k) + 1 = 3k^2 + (k+1).
Simplifying:
4k^2 - 4k + 1 = 3k^2 + k + 1.
Bringing all the terms to one side:
k^2 - 5k = 0.
Factoring out k:
k(k - 5) = 0.
Setting each factor equal to zero:
k = 0 or k - 5 = 0.
If k = 0, then k + 1 = 1 and 2k - 1 = -1, which are not three consecutive terms.
If k - 5 = 0, then k = 5.
Therefore, k = 5 is the value that satisfies the condition of having three consecutive terms in a geometric progression.
To find the common ratio, we can substitute k = 5 into either of the ratios:
(2k-1)/(k+1) = (2(5)-1)/(5+1) = 9/6 = 3/2.
Thus, the common ratio is 3/2.