Find the G.P in which 4th term is 3 and the 7th term is 8/9.

4th term is 3 ---> ar^3 = 3

7th term is 8/9 --> ar^6 = 8/9

divide them:
r^3 = 8/27
r = 2/3

in ar^3 = 3
a(8/27) = 3
a = 81/8

GP is 81/8 , 27/4 , 9/2 , 3 , .....

Thnku so much

Why did the fourth term of the geometric progression become a rebel? Because it wanted to break free from boring patterns and be a 3! And then the seventh term jumped in saying, "Hey, I want to be a fraction of fun too!" Imagine its surprise when it ended up as 8/9. So, the rebellious GP goes like this: 1, 1/3, 1/9, 3, 1/27, 1/81, 8/9... It's a crazy pattern, but hey, it's what the terms wanted!

To find the geometric progression (G.P), we need to determine the common ratio (r) first.

Let the first term be "a" and the common ratio be "r".

The formula for the nth term of a G.P is given by:
tn = a * r^(n-1)

Given the 4th term is 3, we can write the equation as:
3 = a * r^3 ..........(Equation 1)

And given the 7th term is 8/9, we can write the equation as:
8/9 = a * r^6 ..........(Equation 2)

We can divide Equation 2 by Equation 1 to eliminate "a":
(8/9) / 3 = (a * r^6) / (a * r^3)
8/27 = r^6 / r^3
8/27 = r^(6-3)
8/27 = r^3

Taking the cube root of both sides of the equation, we get:
(r^3)^(1/3) = (8/27)^(1/3)
r = 2/3

Now, we substitute the value of "r" in Equation 1 to find "a":
3 = a * (2/3)^3
3 = a * 8/27

Multiplying both sides of the equation by 27/8, we get:
27/8 * 3 = a
81/8 = a

Therefore, the geometric progression (G.P) is:
a, ar, ar^2, ar^3, ar^4, ar^5, ...

Substituting the values, we have:
81/8, (2/3) * (81/8), (2/3)^2 * (81/8), (2/3)^3 * (81/8), ...

Simplifying, we get:
81/8, 54/8, 36/8, 24/8, ...

To find the geometric progression (G.P), we need to determine the common ratio (r).

Let's denote the first term of the G.P as 'a' and the common ratio as 'r'.

We are given that the 4th term is 3, which means:
a * r^3 = 3 -- Equation 1

Similarly, the 7th term is 8/9, so:
a * r^6 = 8/9 -- Equation 2

To find the values of 'a' and 'r', we need to solve these two equations simultaneously.

Let's manipulate Equation 1 to solve for 'a':
a = 3 / r^3

Substituting this value of 'a' into Equation 2, we get:
(3 / r^3) * r^6 = 8/9
3 * r^(6-3) = 8/9
3 * r^3 = 8/9

Now, let's solve this equation for 'r':

r^3 = (8/9) / 3
r^3 = 8/27

Taking the cube root of both sides, we find:
r = (cube root of 8) / (cube root of 27)
r = 2/3

Now, substitute this value of 'r' into Equation 1 to find 'a':
a * (2/3)^3 = 3
a * 8/27 = 3
a = (3 * 27) / 8
a = 81/8

So, the first term 'a' is 81/8 and the common ratio 'r' is 2/3. Therefore, the G.P we are looking for is:
81/8, 27/8, 9/8, 3/8, 1/8, 1/27, 1/81, ...