Solving a Geometric Progression Problem

This problem asks us to identify a specific Geometric Progression (GP) based on two given pieces of information: its second term and the sum of its infinite terms.

A Geometric Progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The general form of a GP is \(a, ar, ar^2, ar^3, \ldots\), where \(a\) is the first term and \(r\) is the common ratio.

Understanding the Given Information

We are given:

• The second term of the GP is 2. In the general form, the second term is \(ar\). So, we have the equation:
  ar = 2 \quad \ldots(1)
• The sum of its infinite terms is 8. The formula for the sum of an infinite GP is S_{\infty} = \frac{a}{1-r}, provided that the absolute value of the common ratio \left|r\right| < 1. So, we have the equation:
  \frac{a}{1-r} = 8 \quad \ldots(2)

Solving for the First Term and Common Ratio

We now have a system of two equations with two variables, \(a\) and \(r\).

From Equation (1), we can express \(a\) in terms of \(r\):

a = \frac{2}{r}

Now, substitute this expression for \(a\) into Equation (2):

\frac{\left(\frac{2}{r}\right)}{1-r} = 8

Simplify the left side:

\frac{2}{r(1-r)} = 8

Multiply both sides by r(1-r):

2 = 8r(1-r)

Divide both sides by 2:

1 = 4r(1-r)

Expand the right side:

1 = 4r - 4r^2

Rearrange the terms to form a quadratic equation:

4r^2 - 4r + 1 = 0

This quadratic equation is a perfect square trinomial:

(2r - 1)^2 = 0

Taking the square root of both sides:

2r - 1 = 0

Solve for \(r\):

2r = 1

r = \frac{1}{2}

We must check if the condition \left|r\right| < 1 is met. \left|\frac{1}{2}\right| = \frac{1}{2}, which is less than 1. So, this value of \(r\) is valid for the infinite sum formula.

Now, substitute the value of \(r = \frac{1}{2}\) back into Equation (1) to find \(a\):

a \left(\frac{1}{2}\right) = 2

Multiply both sides by 2:

a = 4

So, the first term is 4 and the common ratio is 1/2.

Constructing the Geometric Progression

Using the first term \(a=4\) and the common ratio \(r=\frac{1}{2}\), the terms of the GP are:

• 1st term: a = 4
• 2nd term: ar = 4 \times \frac{1}{2} = 2 (Matches the given information)
• 3rd term: ar^2 = 4 \times \left(\frac{1}{2}\right)^2 = 4 \times \frac{1}{4} = 1
• 4th term: ar^3 = 4 \times \left(\frac{1}{2}\right)^3 = 4 \times \frac{1}{8} = \frac{1}{2}
• 5th term: ar^4 = 4 \times \left(\frac{1}{2}\right)^4 = 4 \times \frac{1}{16} = \frac{1}{4} = \frac{1}{2^2}

The Geometric Progression is 4, 2, 1, \frac{1}{2}, \frac{1}{4}, \ldots.

Comparing with the Options

Let's check which option matches the GP we found:

Option 3 matches the Geometric Progression we found with first term \(a=4\) and common ratio \(r=\frac{1}{2}\).

Conclusion

The Geometric Progression whose second term is 2 and the sum of its infinite terms is 8 is 4, 2, 1, \frac{1}{2}, \frac{1}{4}, \ldots.

Geometric Progression Revision Table

Additional Information on Infinite Geometric Series

An infinite geometric series is the sum of an infinite number of terms in a Geometric Progression: a + ar + ar^2 + ar^3 + \ldots. The sum of an infinite geometric series converges (approaches a finite value) only if the absolute value of the common ratio \left|r\right| is less than 1 (-1 < r < 1). If \left|r\right| \geq 1, the terms do not approach zero, and the sum diverges (does not approach a finite value).

In this problem, we found r = \frac{1}{2}, and \left|\frac{1}{2}\right| = \frac{1}{2} < 1, which confirms that the infinite sum exists and the formula S_{\infty} = \frac{a}{1-r} is applicable.