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Question: If \[a,\,b,\,c\] are real numbers and \[{a^2},{b^2},{c^2}\] are in AP, then which of the following c...

If a, b, ca,\,b,\,c are real numbers and a2,b2,c2{a^2},{b^2},{c^2} are in AP, then which of the following combinations are in APAP ?

Explanation

Solution

In this, arithmetic progression is between a2,b2,c2{a^2},{b^2},{c^2}. So, the difference between the AP is equal to the succession between the given. So, we should make an equation out of this given arithmetic progression and then we will get the following answer.

Complete step by step answer:
As according to the question a2,b2,c2{a^2},{b^2},{c^2}are in the AP, we have:
b2−a2=c2−b2{b^2} - {a^2} = {c^2} - {b^2}
When we expand it according to the basic formula of algebra, we get:
(b−a)(b+a)=(c−b)(c+b)\left( {b - a} \right)\left( {b + a} \right) = \left( {c - b} \right)\left( {c + b} \right)
Now, we will cross multiply them, and we get:
(b−ac+b)=(c−bb+a)\left( {\dfrac{{b - a}}{{c + b}}} \right) = \left( {\dfrac{{c - b}}{{b + a}}} \right)
Now, we will try to divide both the sides with (c+a)(c + a), and we get:
(b+c−c−a(c+b)(c+a))=(c+a−a−b(b+a)(c+a))\left( {\dfrac{{b + c - c - a}}{{\left( {c + b} \right)\left( {c + a} \right)}}} \right) = \left( {\dfrac{{c + a - a - b}}{{\left( {b + a} \right)\left( {c + a} \right)}}} \right)
⇒(1c+a)−(1b+c)=(1a+b)−(1c+a)\Rightarrow \left( {\dfrac{1}{{c + a}}} \right) - \left( {\dfrac{1}{{b + c}}} \right) = \left( {\dfrac{1}{{a + b}}} \right) - \left( {\dfrac{1}{{c + a}}} \right)

So, we can tell that:
(1b+c),(1c+a),(1a+b)\left( {\dfrac{1}{{b + c}}} \right),\left( {\dfrac{1}{{c + a}}} \right),\left( {\dfrac{1}{{a + b}}} \right) are in arithmetic progression
We can also write it as:
(a+b+cb+c),(a+b+cc+a),(a+b+ca+b)\left( {\dfrac{{a + b + c}}{{b + c}}} \right),\left( {\dfrac{{a + b + c}}{{c + a}}} \right),\left( {\dfrac{{a + b + c}}{{a + b}}} \right)
⇒(ab+c+1),(bc+a+1),(ca+b+1)\Rightarrow \left( {\dfrac{a}{{b + c}} + 1} \right),\left( {\dfrac{b}{{c + a}} + 1} \right),\left( {\dfrac{c}{{a + b}} + 1} \right)
∴(ab+c),(bc+a),(ca+b)\therefore \left( {\dfrac{a}{{b + c}}} \right),\left( {\dfrac{b}{{c + a}}} \right),\left( {\dfrac{c}{{a + b}}} \right)
These are also in arithmetic progression.

Note: Arithmetic progression can be any sequence of numbers in which the difference of any number should be a constant value. It can be understood by an example, the sequence of numbers 1,2,3,4,5,6…is an AP, in which we can see a common difference between the numbers. If we take two numbers like 2 and 3 the difference between 2 and 3 is 1, and the difference between the ap is 1.