If k+1,2k-1 and 3k+1 are three consecutive terms of a GP, find the common ratio if the final answer is 8
We know that in a geometric progression, each term is obtained by multiplying the previous term by a common ratio. Let's express the terms in terms of the common ratio:
k + 1 = a
2k - 1 = ar
3k + 1 = ar^2
Now, we can manipulate these equations to eliminate k:
2(k + 1) - (2k - 1) = 3k + 1
2k + 2 - 2k + 1 = 3k + 1
3 = 3k + 1
3k = 2
k = 2/3
Now that we have k, we can substitute it back into any of the equations to find the values of the terms. Let's use the first equation:
k + 1 = a
2/3 + 1 = a
5/3 = a
So, the first term of the geometric progression is 5/3.
Now, we can find the common ratio:
2k - 1 = ar
2(2/3) - 1 = (5/3) * r
4/3 - 1 = 5r/3
-3/3 = (5r - 3)/3
-1 = 5r - 3
5r = -1 + 3
5r = 2
r = 2/5
Therefore, the common ratio is 2/5.