If 4/3,m,n,1 form a g.p the product of m and n is

Answer 1

We know that for a geometric progression (g.p.), the ratio between consecutive terms is constant. Let's call this ratio "r". Then:

m/4/3 = r (equation 1)
n/m = r (equation 2)
1/n = r (equation 3)

We can simplify equation 3 to get:

n = 1/r

Substituting this into equation 2, we get:

1/(mr) = r

Multiplying both sides by mr, we get:

1 = r^2m

Substituting equation 1 into this, we get:

1 = r^(2)(4/3)

Simplifying this, we get:

r^2 = 3/4

Taking the square root of both sides (remembering that r must be positive for a g.p.), we get:

r = sqrt(3)/2

Substituting this into equation 2, we get:

n/m = sqrt(3)/2

Multiplying both sides by m, we get:

n = m(sqrt(3)/2)

Finally, we know that the product of m and n is:

mn = m^2(sqrt(3)/2)

We can't determine the exact value of mn without knowing the value of m, but we can say that it is equal to m^2 times a constant.

Answer 2

Another math problem botched by the AI ....

Easy way:

a = 4/3
m = ar
n = ar^2
1 = ar^3 --> r^3 = 1/a

mn = (ar)(ar^2) = a^2 r^3
= (16/9)(1/a)
= (16/9)(3/4) = 4/3

or, the long way ...

a = 4/3
we are also given term 4,
ar^3 = 1
(4/3)r^3 = 1
r^3 = 3/4 or 6/8
r = cuberoot(6)/2 = appr .90856

so m = ar = (4/3)(cuberoot(6)/2) = (2/3) cuberoot(6) or (2/3)6^(1/3)
= appr 1.211414
n = term(2)*r = (1/3)( cuberoot(6) )^2 or (1/3)6^(1/3)
= appr 1.10064

mn = (2/3)6^(1/3)*(1/3)6^(1/3)
= (2/9) 6^(2/3)
= 1.211414 * 1.10064 = 4/3

verified by calculator

Answer 3

I apologize for the mistake in my previous response. Thank you for providing the correct solution.