if 4/3, m, 1, n...form a GP, the product of m and n is
Common ratio is 1/m which is equal to n/1 (i.e taken from any two close terms).
Thus: 1/m=n/1
Cross multiply
mn=1
Therefore the product of m and n is 1.
T3=ar² ÷ m(T1)=ar
1=(4/3)r²
÷
m=(4/3)r
Therefore m=1/r
For n(T4),
T4= (3/4)×r³
÷
(T3)1= (3/4)r²
Therefore n/1=r
m=1/r
×
n=r
Therefore, m×n=1/r×r
MN=1
The answer is 1
the term between them
To determine the product of "m" and "n" in the given geometric progression (GP), we need to find the common ratio (r).
In a GP, each term (except the first term) is obtained by multiplying the previous term by a fixed value called the common ratio. Let's find the common ratio by comparing the ratios of consecutive terms:
Term 2 / Term 1 = m / (4/3)
Term 3 / Term 2 = 1 / m
Term 4 / Term 3 = n / 1
Since the terms form a geometric progression, these ratios should be equal. Setting up equations:
m / (4/3) = 1 / m --> Equation 1
1 / m = n / 1 --> Equation 2
Simplifying Equation 2, we get:
n = m^2 --> Equation 2'
Now, substitute Equation 2' into Equation 1 to solve for "m":
m / (4/3) = 1 / m
m^2 = 4/3
Taking the square root of both sides:
m = ± √(4/3)
Since m is positive (as a negative value would result in a negative product), we take the positive square root:
m = √(4/3) = √4 / √3 = 2 / √3 = (2√3) / 3
To find "n", substitute the value of "m" into Equation 2':
n = (2√3 / 3)^2
n = (4 * 3) / 9
n = 12 / 9
n = 4 / 3
Therefore, the product of "m" and "n" is:
m * n = [(2√3) / 3] * [4 / 3]
= (8√3) / 9
sorry...forgot the square
the square of the term between them