In this problem, we are given an infinite series in geometric progression (GP) and we are asked to find the sum of some of its terms and the sum of their squares.



Let the first term of the GP be a and the common ratio be r. Then, the infinite series can be written as:

a + ar + ar^2 + ar^3 + ...

Using the formula for the sum of an infinite GP, we get:

sum = a / (1 - r)

Multiplying both sides by r, we get:

r*sum = ar / (1 - r)

Subtracting the second equation from the first, we get:

sum - r*sum = a / (1 - r) - ar / (1 - r)

Simplifying, we get:

sum = a / (1 - r)

Similarly, we can find the sum of the squares of the terms of the GP as:

sum of squares = a^2 / (1 - r^2)

Now, we are given that the sum of some of the terms of the GP is 2 and the sum of their squares is 4/3. Let the sum of these terms be S and the sum of their squares be T. Then, we have:

S = a(1 - r^n) / (1 - r)

T = a^2(r^(2n) - 1) / (r^2 - 1)

where n is the number of terms being summed.

We can solve these equations for a and r in terms of S and T. First, we isolate a in the equation for S:

a = S(1 - r) / (1 - r^n)

Substituting this into the equation for T, we get:

T = S^2(1 - r^(2n)) / (1 - r^n)^2 * (r^2 - 1)

Simplifying, we get:

r^(2n) = (T(1 - r^n)^2 * (r^2 - 1)) / (S^2(1 - r^(2n)))

This equation can be solved for r^(2n) using numerical methods. Once we have r^(2n), we can find r using the equation:

r = (r^(2n))^(1/2n)

Finally, we can substitute a and r into the equation for the sum of the GP to find the first term a:

sum = a / (1 - r)

Once we have a, we can find the common ratio r using the equation:

r = (S - a) / (S - 2a)

Finally, we can write out the infinite series:

a + ar + ar^2 + ar^3 + ...



In conclusion, we can find the sum of an infinite series in GP given its first term and common ratio. We can also use the sum and sum of squares of a subset of terms to solve for the first term and common ratio.