Geometric Progression: Ratio When Infinite Sum Is 'm' Times N Terms

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Hey guys! Let's dive into an interesting problem involving geometric progressions (GPs). We're going to figure out the common ratio of a GP where the sum to infinity is a multiple ('m' times) of the sum of its first 'n' terms. Sounds intriguing, right? Buckle up, and let’s get started!

Understanding Geometric Progressions

First off, let's quickly recap what a geometric progression actually is. A geometric progression, or GP, is simply a sequence of numbers where each term is found by multiplying the previous term by a constant value. This constant value is what we call the common ratio, often denoted by 'r'. For example, the sequence 2, 4, 8, 16,... is a GP where the common ratio is 2 (each term is twice the previous one).

Key elements of a GP that we need to remember are:

  • First term (a): This is the starting number of the sequence.
  • Common ratio (r): The factor by which each term is multiplied to get the next term.
  • nth term (Tn): The term at the nth position in the sequence, given by the formula T_n = a * r^(n-1).
  • Sum of the first n terms (Sn): The sum of the first n terms of the GP, calculated using the formula S_n = a * (1 - r^n) / (1 - r) when r < 1 and S_n = a * (r^n - 1) / (r - 1) when r > 1.
  • Sum to infinity (S∞): When the absolute value of the common ratio |r| is less than 1, the GP converges, and we can calculate the sum of an infinite number of terms using the formula S_∞ = a / (1 - r).

Sum to Infinity: A Quick Dive

The concept of sum to infinity might seem a bit mind-bending at first. How can we add up an infinite number of terms and get a finite result? Well, it works for GPs where the terms keep getting smaller and smaller as we go further down the sequence. Imagine a pizza; you eat half, then half of the remaining half, then half of that, and so on. You're adding smaller and smaller fractions, and eventually, the pieces become so tiny that they don't significantly add to the total. This is the basic idea behind the sum to infinity.

Mathematically, the formula for the sum to infinity (S∞) of a GP is:
S_∞ = a / (1 - r),
where 'a' is the first term and 'r' is the common ratio, with the condition that |r| < 1. This condition is crucial because if |r| is greater than or equal to 1, the terms either stay the same size or get larger, and the sum will tend towards infinity (or negative infinity).

Problem Statement: Decoding the Question

Now that we've refreshed our understanding of GPs, let's break down the problem statement. We're given a geometric progression with an infinite number of terms, and it's a decreasing GP (meaning the terms are getting smaller). We're told that the sum of all these infinite terms (S∞) is 'm' times the sum of the first 'n' terms (Sn). Our mission is to find the common ratio 'r'.

So, in mathematical terms, we have:

S_∞ = m * S_n

Our goal is to express both S∞ and Sn in terms of 'a' and 'r', and then solve for 'r'. This involves applying the formulas we discussed earlier and doing a bit of algebraic manipulation.

Setting up the Equations

Okay, let’s put the formulas into action. We know:

  • Sum to infinity (S∞) = a / (1 - r)
  • Sum of the first n terms (Sn) = a * (1 - r^n) / (1 - r)

Now, we substitute these expressions into our given equation:

a / (1 - r) = m * [a * (1 - r^n) / (1 - r)]

This equation is the heart of our problem. It relates the sum to infinity to the sum of the first 'n' terms, and it's where we'll start our algebraic journey to find 'r'. The next step is to simplify this equation, which will make it easier to isolate and solve for the common ratio.

Solving for the Common Ratio (r)

Let's simplify the equation we derived in the previous section:

a / (1 - r) = m * [a * (1 - r^n) / (1 - r)]

Notice that we have 'a / (1 - r)' on both sides of the equation. We can divide both sides by 'a / (1 - r)' (assuming 'a' is not zero and 'r' is not 1, which are standard assumptions for GP problems). This gives us:

1 = m * (1 - r^n)

Now, let’s isolate the term with 'r':

1 / m = 1 - r^n r^n = 1 - (1 / m) r^n = (m - 1) / m

Great! We've got r^n isolated. To find 'r', we need to take the nth root of both sides:

r = [(m - 1) / m]^(1/n)

And there we have it! We've found the common ratio 'r' in terms of 'm' and 'n'.

Interpreting the Solution

The solution, r = [(m - 1) / m]^(1/n), tells us that the common ratio depends on both m, which is the factor by which the sum to infinity exceeds the sum of the first n terms, and n, the number of terms considered in the initial sum. Let's think about what this means:

  • m > 1: Since the sum to infinity is m times the sum of the first n terms, m must be greater than 1. If m were equal to 1, the sum to infinity would be equal to the sum of the first n terms, which isn't generally true for an infinite GP. If m were less than 1, it would imply that the sum to infinity is smaller than the sum of the first n terms, which is impossible for a decreasing GP with all positive terms.

  • (m - 1) / m: This fraction will always be between 0 and 1 when m > 1. As m gets larger, this fraction approaches 1. This makes sense because if the sum to infinity is a very large multiple of the sum of the first n terms, then the terms beyond the nth term contribute a significant amount to the sum, implying that the terms are decreasing relatively slowly. This means the common ratio should be closer to 1.

  • Taking the nth root: The exponent (1/n) has the effect of scaling the value of [( m - 1) / m ]. As n increases, the nth root will tend to 1. This is because a larger n means we are considering the effect of the ratio over more terms, and the overall decrease to reach the sum to infinity needs to be less drastic per term.

Example Scenario

To make this solution more concrete, let’s consider a quick example. Suppose the sum to infinity of a GP is 2 times the sum of its first 3 terms (so, m = 2 and n = 3). What would be the common ratio?

Using our formula:

r = [(2 - 1) / 2]^(1/3) r = (1/2)^(1/3) r ≈ 0.7937

So, the common ratio would be approximately 0.7937. This means each term in the GP is roughly 79.37% of the previous term, making it a decreasing GP as expected.

Key Takeaways

Let's summarise the key things we've learned in this article:

  1. Geometric Progressions: We revisited the fundamentals of GPs, including the definitions of the first term, common ratio, nth term, sum of the first n terms, and the sum to infinity.
  2. Sum to Infinity: We understood the concept of the sum to infinity and its formula, S∞ = a / (1 - r), which is applicable when |r| < 1.
  3. Problem Solving: We successfully set up and solved an equation relating the sum to infinity to the sum of the first n terms, and we found the common ratio r = [(m - 1) / m]^(1/n).
  4. Interpretation: We discussed the implications of the solution and how the common ratio depends on m and n.

I hope this article has helped you understand how to find the common ratio of a GP when given the relationship between the sum to infinity and the sum of its first n terms. Remember, the key is to break down the problem into manageable steps, use the correct formulas, and carefully simplify the equations. Keep practicing, and you'll become a GP whiz in no time!