Finding the Number of Arithmetic Means in a Sequence with a Given Ratio
In this article, we will explore a specific problem involving arithmetic means inserted between two numbers in an arithmetic progression (AP). We will delve into the steps required to find the value of n, the number of inserted means, given the ratio of the first to the last mean. This is a classic problem in the realm of arithmetic sequences and progressions, and understanding it can be valuable for students and professionals alike.
Problem Statement
Consider an arithmetic progression (AP) where 20 arithmetic means are inserted between 20 and 80 such that the ratio of the first mean to the last mean is 1:3. We need to find the value of n.
Step-by-Step Solution
Let's start by understanding the basics. An arithmetic progression is a sequence of numbers in which the difference between any two successive members is constant. This constant difference is referred to as the common difference, denoted as d.
Formulating the Problem
Given:
The first term of the AP, T1, is 20. The last term of the AP, Tn 1, is 80. The number of arithmetic means inserted between 20 and 80 is n. The ratio of the first mean to the last mean is 1:3.Let's denote the terms of the AP as T1, T2, T3, ..., Tn 2. The first term is given as 20, and the last term is given as 80. Since there are n means between 20 and 80, the total number of terms in the AP is n 2.
Equations in Action
The general formula for the k-th term in an arithmetic progression is:
Tk T1 (k-1) d
Using the given terms:
Tn 2 T1 n d 20 n d 80 Tn 2 - T1 (n-1)d 60From the equation (n-1)d 60:
d 60 / (n-1)
Finding the Means and the Ratio
The first mean in the arithmetic progression is:
T2 T1 d 20 d
The last mean in the arithmetic progression is:
Tn 1 Tn 2 - d 80 - d
The given ratio of the first mean to the last mean is 1:3:
T2 / Tn 1 1 / 3
Substituting the expressions for T2 and Tn 1:
(20 d) / (80 - d) 1 / 3
Cross-multiplying to solve for d:
3(20 d) 80 - d
60 3d 80 - d
4d 20
d 5
Substitute d back into the equation (n-1)d 60:
(n-1) * 5 60
n-1 60 / 5
n-1 12
n 13
Verification
To verify the solution, we can check the values of the terms in the arithmetic progression:
T1 20
T14 20 12 * 5 20 60 80
The values are correct, confirming that the value of n is indeed 13.
Conclusion
By solving the arithmetic progression problem step-by-step, we have successfully determined that the number of arithmetic means inserted between 20 and 80, given the specified ratio, is n 13. This problem not only tests the understanding of arithmetic sequences but also reinforces the application of algebraic manipulation and solving equations.