If 1/3 ,m,1,n form a GP the product of m and n
To form a geometric progression (GP), we know that the ratio between any two consecutive terms must be the same.
Let's consider the given terms: 1/3, m, 1, n.
The ratio between the first and second terms is: m / (1/3) = 3m.
The ratio between the second and third terms is: 1 / m.
The ratio between the third and fourth terms is: n / 1 = n.
Since the ratio between the second and third terms is 1/m and the ratio between the third and fourth terms is n, we can equate them:
1/m = n
Solving for m, we get:
m = 1/n
Therefore, the product of m and n is:
m * n = (1/n) * n = 1
or, since 1 is the geometric mean of m and n,
√(mn) = 1
so mn = 1
Yes, you are correct. Since 1 is the geometric mean of m and n, we can also find the product of m and n by taking the square root of their product and setting it equal to 1.
√(mn) = 1
Squaring both sides:
mn = 1
Therefore, the product of m and n is indeed 1.
To determine the product of m and n in the geometric progression (GP) formed by 1/3, m, 1, and n, we can use the formula for a GP.
In a GP, each term is obtained by multiplying the previous term by a constant called the common ratio (r). In this case, we have 1/3, m, 1, and n.
So, we can write the relationship between these terms as follows:
m = (1/3) * r
1 = m * r
n = 1 * r
To find the product of m and n, we need to eliminate the common ratio, r. We can do this by dividing the second equation by the first:
1/3 * r / (1/3) = (m * r) / (1/3)
r = 3 * m
Now, substitute this value of r into the third equation:
n = 1 * (3 * m)
n = 3 * m
Therefore, the product of m and n is (3 * m) * (3 * m), which simplifies to 9 * m^2.