k+1,2k-1,3k+1 are three consecutive terms of a geometric progression, find the possible values of the common ratio
K=8
If k+1, 2k-1 and 3k+1 are three consecutive terms of a gp. Find the possible value of the common ratio
K=8
is k+1 ratio
1,5 and 6
NO
To find the possible values of the common ratio, we need to determine if these three terms form a geometric progression. In a geometric progression, each term is obtained by multiplying the previous term by a constant value known as the common ratio.
Let's write these three terms as an equation:
k + 1 = a (first term)
2k - 1 = ar (second term)
3k + 1 = ar^2 (third term)
Now, we can solve for the common ratio (r).
From the given equation, we have:
2k - 1 = ar
Rearranging, we get:
r = (2k - 1) / a
Similarly, for the third term, we have:
3k + 1 = ar^2
Rearranging, we get:
r^2 = (3k + 1) / a
Now, let's equate the two values of r we found:
(2k - 1) / a = (3k + 1) / a^2
Cross-multiplying, we have:
a(2k - 1) = (3k + 1)
Expanding, we get:
2ak - a = 3k + 1
Grouping the similar terms, we have:
(2a - 3)k = a + 1
Now, there can be two cases:
Case 1: If 2a - 3 = 0, then the equation becomes:
0k = a + 1
This implies that a + 1 = 0. Therefore, a = -1. However, this value of a leads to division by zero, which is not valid.
Case 2: If 2a - 3 ≠ 0, then we can divide both sides of the equation by (2a - 3):
k = (a + 1) / (2a - 3)
Therefore, the possible values of the common ratio (r) are all the values of k that satisfy the equation k = (a + 1) / (2a - 3), where 2a - 3 ≠ 0.
So, to find the possible values of the common ratio, we need to determine the values of k that satisfy the equation k = (a + 1) / (2a - 3), where 2a - 3 ≠ 0.