If k+1,2k-1and 3k+1 are three consecutive terms of a GP ,find the value of the common ratio

Answer 1

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.