Geometric Progression (GP)

A Geometric Progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.

The formula for the nth term of a GP is given by:
\(a_n = a_1 \times r^{(n-1)}\)
where \(a_n\) is the nth term, \(a_1\) is the first term, and \(r\) is the common ratio.

Sum of an Infinite Geometric Progression

The sum of an infinite GP can be calculated using the formula:
\(S = \frac{a_1}{1 - r}\)
where \(S\) is the sum, \(a_1\) is the first term, and \(r\) is the common ratio.

Sum of the Squares of an Infinite Geometric Progression

The sum of the squares of an infinite GP can be calculated using the formula:
\(S^2 = \frac{a_1^2}{(1 - r)^2}\) + \(\frac{a_1^2 \times r^2}{(1 - r)^4}\)
where \(S\) is the sum, \(a_1\) is the first term, and \(r\) is the common ratio.

Given Information

In this problem, we are given that the sum of the infinite GP is 2 and the sum of the squares of the infinite GP is 4/3.

So, we can write the equations as:
\(2 = \frac{a_1}{1 - r}\)
\(4/3 = \frac{a_1^2}{(1 - r)^2}\) + \(\frac{a_1^2 \times r^2}{(1 - r)^4}\)

Calculating the First Term

To find the first term, we need to solve the above equations simultaneously.

From the first equation, we can express \(a_1\) in terms of \(r\) as:
\(a_1 = 2(1 - r)\)

Substituting this value of \(a_1\) in the second equation, we get:
\(\frac{4}{3} = \frac{(2(1 - r))^2}{(1 - r)^2}\) + \(\frac{(2(1 - r))^2 \times r^2}{(1 - r)^4}\)

Simplifying the equation, we get:
\(\frac{4}{3} = \frac{4(1 - r)^2}{(1 - r)^2}\) + \(\frac{4(1 - r)^2 \times r^2}{(1 - r)^4}\)

Cancelling out the common terms, we get:
\(\frac{4}{3} = 4 + 4r^2\)

Simplifying further, we have:
\(r^2 + 1 = \frac{4}{3}\)

Solving this quadratic equation, we get two solutions for \(r\):
\(r = \pm \frac{1}{\sqrt{3}}\)

Choosing the Appropriate Solution

Since the common ratio of a GP must be a non-zero number, we can