Problem about arithmetic progression

In summary, the conversation is about a math problem where the goal is to prove that if a^2, b^2, and c^2 are in arithmetic progression, then 1 / (b + c), 1 / (c + a), and 1 / (a + b) are also in arithmetic progression. The participants discuss the given assumptions and try to find the solution, eventually concluding that the problem can be solved by assuming m = n. The conversation ends with one participant sharing a solution from another forum.
  • #1
Augustus58
2
0
Hi,

can't solve following prob:
Let a, b and c be real numbers.
Given that a^2, b^2 and c^2 are in arithmetic progression show that 1 / (b + c), 1 / (c + a) and 1 / (a + b) are also in arithmetic progression.

From assumptions: b^2 = a^2 + nk and c^2 = b^2 + mk where k is some real number and n and m are whole numbers.

Then f.e. b^2 - a^2 = nk and 1 / (a + b) = (a - b) / (a^2 - b^2) =(b - a) / (nk).
Further 1 / (c + a) = (c - a) / ((k(m + n)) and 1 / (b + c) = (c - b) / (mk).

I tried to get something out from 1 / (b + c) - 1 / (c + a) and 1 / (c + a) - 1 / (a + b) related to definition of AP (f.e. pr and qr where p and q are whole numbers and r is real number).

Not solved yet :) Seems easy.. x)
 
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  • #2
Welcome to PF!

This looks like a homework problem. PF uses a homework template where you write down the problem, state relevant formulas and then show some work. We can't help you much without some work shown.

To start things off:

Have you established that it is true?

Do you have a simple progression that illustrates the theorem ie values for a,b,c and m,n and k?
 
  • #3
I wonder if it was meant that the numbers are meant to be consecutive members of an arithmetic progression, so m = b.

Work out the differences between 1 / (b + c) and 1 / (c + a) and also 1 / (c + a) and 1/(a+b). Make sure the fractions have the same divisors.
 
  • #5


I would first try to understand the problem and the given information. In this case, we are given three real numbers a, b, and c, and we are told that their squares are in arithmetic progression. This means that there is a common difference between each of their squares. We are then asked to show that the reciprocals of the sums of these numbers (1/(b+c), 1/(c+a), and 1/(a+b)) are also in arithmetic progression.

To solve this problem, we can start by using the given information to find relationships between the numbers. For example, we know that b^2 = a^2 + nk and c^2 = b^2 + mk, where k is a real number and n and m are whole numbers. We can then substitute these expressions into the reciprocals of the sums: 1/(b+c) = 1/((a+b)+c) = 1/(a+b) * 1/(1+c/(a+b)) = 1/(a+b) * 1/(1+(c^2-a^2)/(a^2+b^2)) = 1/(a+b) * 1/(1+(mk)/(nk)) = 1/(a+b) * 1/(1+m/n) = 1/(a+b) * (n/(n+m)). Similarly, we can find expressions for 1/(c+a) and 1/(a+b).

Next, we can use the definition of arithmetic progression (AP) to show that these expressions are also in AP. The definition of AP states that the difference between any two consecutive terms is constant. Therefore, if we take the difference between 1/(b+c) and 1/(c+a), we get 1/(b+c) - 1/(c+a) = 1/(a+b) * (n/(n+m)) - 1/(a+b) * (m/(n+m)) = 1/(a+b) * (n-m)/(n+m). This is a constant value, since n and m are whole numbers. Similarly, the difference between 1/(c+a) and 1/(a+b) is also a constant value.

Therefore, we have shown that the reciprocals of the sums of a, b, and c are in arithmetic progression. This is just one possible approach to solving this problem, and there may be
 

1. What is an arithmetic progression?

An arithmetic progression is a sequence of numbers where the difference between consecutive terms is constant. It can be represented as a linear function in the form of f(x) = ax + b, where a is the common difference and b is the initial term.

2. How do you find the common difference in an arithmetic progression?

To find the common difference in an arithmetic progression, subtract any term from its previous term. The result will be the common difference. For example, in the sequence 2, 5, 8, 11, 14, the common difference is 3 because 5-2=3 and 8-5=3, and so on.

3. Can an arithmetic progression have a negative common difference?

Yes, an arithmetic progression can have a negative common difference. This means that the terms in the sequence will decrease with each consecutive term instead of increasing. For example, in the sequence 10, 7, 4, 1, -2, the common difference is -3.

4. What is the formula for finding the nth term in an arithmetic progression?

The formula for finding the nth term in an arithmetic progression is an = a1 + (n-1)d, where a1 is the initial term, d is the common difference, and n is the term number. For example, in the sequence 3, 7, 11, 15, 19, the 5th term is found by plugging in a1 = 3, d = 4, and n = 5, giving us a5 = 3 + (5-1)4 = 19.

5. How is an arithmetic progression used in real-world applications?

Arithmetic progressions are commonly used in financial calculations, such as calculating compound interest or depreciation. They are also used in physics and engineering to model linear relationships, such as velocity or acceleration over time. In computer science, arithmetic progressions are used in algorithms and data structures for efficient storage and retrieval of data.

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