Solveeit Logo
Login

Question

Question: If \(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\) are in A.P, prove that \(\dfrac{{\left( {b + c} \right...

If 1a,1b,1c\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c} are in A.P, prove that (b+c)a,(c+a)b,(a+b)c\dfrac{{\left( {b + c} \right)}}{a},\dfrac{{\left( {c + a} \right)}}{b},\dfrac{{\left( {a + b} \right)}}{c} are in A.P.

Explanation

Solution

It is given in the question that if 1a,1b,1c\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c} are in A.P.
Then, we have to prove that (b+c)a,(c+a)b,(a+b)c\dfrac{{\left( {b + c} \right)}}{a},\dfrac{{\left( {c + a} \right)}}{b},\dfrac{{\left( {a + b} \right)}}{c} are in A.P.
First, we will multiply a+b+ca + b + c with 1a,1b,1c\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c} . Then, we will subtract with 1 from the equation which obtained by multiply a+b+ca + b + c with 1a,1b,1c\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}. Finally, we will get the answer.

Complete step by step solution:
It is given in the question that if 1a,1b,1c\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c} are in A.P.
Then, we have to prove that (b+c)a,(c+a)b,(a+b)c\dfrac{{\left( {b + c} \right)}}{a},\dfrac{{\left( {c + a} \right)}}{b},\dfrac{{\left( {a + b} \right)}}{c} are in A.P.
Since, we know that if we multiply or divide the numbers which are in AP with a particular number, then the number remains in AP.
Now, multiply a+b+ca + b + c with 1a,1b,1c\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c} , we get,
→a+b+ca,a+b+cb,a+b+cc\to \dfrac{{a + b + c}}{a},\dfrac{{a + b + c}}{b},\dfrac{{a + b + c}}{c} .
Since, if we add or subtract some particular number from the numbers which are in AP, then also the numbers still remain in the AP only.
Now, subtract the above number with 1, we get,
→a+b+ca−1,a+b+cb−1,a+b+cc−1\to \dfrac{{a + b + c}}{a} - 1,\dfrac{{a + b + c}}{b} - 1,\dfrac{{a + b + c}}{c} - 1
→a+b+c−aa,a+b+c−bb,a+b+c−cc\to \dfrac{{a + b + c - a}}{a},\dfrac{{a + b + c - b}}{b},\dfrac{{a + b + c - c}}{c}
→b+ca,a+cb,a+bc\to \dfrac{{b + c}}{a},\dfrac{{a + c}}{b},\dfrac{{a + b}}{c}

Therefore, b+ca,a+cb,a+bc\dfrac{{b + c}}{a},\dfrac{{a + c}}{b},\dfrac{{a + b}}{c} are in A.P.

Note:
Arithmetic Progression: An arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant. Difference here means the second minus first.
If the initial term of an arithmetic progression is a1{a_1} and the common difference of successive member is d, then the nth{n_{th}} term of the sequence is given by,
an=a1+(n−1)d{a_n} = {a_1} + \left( {n - 1} \right)d ,
And the general formula of A.P is,
an=am+(n−m)d{a_n} = {a_m} + \left( {n - m} \right)d.