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Question: Prove that, if \({{a}^{2}},{{b}^{2}},{{c}^{2}}\) are in AP, then the following are also in AP \(\dfr...

Prove that, if a2,b2,c2{{a}^{2}},{{b}^{2}},{{c}^{2}} are in AP, then the following are also in AP ab+c,bc+a,ca+b\dfrac{a}{b+c},\dfrac{b}{c+a},\dfrac{c}{a+b}.

Explanation

Solution

In this question, we are given an AP of three numbers and we have to use it to prove the other three numbers to be in AP. For this, we will take the difference between second and first term of given AP to be equal to difference between third term and second term of given AP. Using that, we will prove our new AP. We will use property as a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right).

Complete step by step answer:
Here, we are given three numbers a2,b2,c2{{a}^{2}},{{b}^{2}},{{c}^{2}} which are in arithmetic progression. We have to prove that ab+c,bc+a,ca+b\dfrac{a}{b+c},\dfrac{b}{c+a},\dfrac{c}{a+b} are also in arithmetic progression. Since, a2,b2,c2{{a}^{2}},{{b}^{2}},{{c}^{2}} are in arithmetic progression, therefore, difference between second term and first term will be equal to difference between third term and second term. Hence, we get b2−a2=c2−b2{{b}^{2}}-{{a}^{2}}={{c}^{2}}-{{b}^{2}}.
As we know, a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) therefore above equation becomes equal to (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, let us divide both sides by (b+a)(c+b)\left( b+a \right)\left( c+b \right) we get:
⇒(b−a)(b+a)(b+a)(c+b)=(c−b)(c+b)(b+a)(c+b)\Rightarrow \dfrac{\left( b-a \right)\left( b+a \right)}{\left( b+a \right)\left( c+b \right)}=\dfrac{\left( c-b \right)\left( c+b \right)}{\left( b+a \right)\left( c+b \right)}
Cancelling (b+a) on left side and (c+b) on right side, we get:
⇒(b−a)(c+b)=(c−b)(b+a)\Rightarrow \dfrac{\left( b-a \right)}{\left( c+b \right)}=\dfrac{\left( c-b \right)}{\left( b+a \right)}
Let us multiply both sides by 1(c+a)\dfrac{1}{\left( c+a \right)} we get:
⇒(b−a)(c+b)(c+a)=(c−b)(b+a)(c+a)\Rightarrow \dfrac{\left( b-a \right)}{\left( c+b \right)\left( c+a \right)}=\dfrac{\left( c-b \right)}{\left( b+a \right)\left( c+a \right)}
For proper symmetry, let us add and subtract c on numerator of left side and add and subtract 'a' on numerator of right side, we get:
⇒(b+c−c−a)(c+b)(c+a)=(c+a−a−b)(b+a)(c+a)\Rightarrow \dfrac{\left( b+c-c-a \right)}{\left( c+b \right)\left( c+a \right)}=\dfrac{\left( c+a-a-b \right)}{\left( b+a \right)\left( c+a \right)}
Taking negative signs common, we get:
⇒(b+c)−(c+a)(c+b)(c+a)=(c+a)−(a+b)(b+a)(c+a)\Rightarrow \dfrac{\left( b+c \right)-\left( c+a \right)}{\left( c+b \right)\left( c+a \right)}=\dfrac{\left( c+a \right)-\left( a+b \right)}{\left( b+a \right)\left( c+a \right)}
Let us now separate terms on both sides, we get:
⇒(b+c)(c+b)(c+a)−(c+a)(c+b)(c+a)=(c+a)(b+a)(c+a)−(a+b)(b+a)(c+a)\Rightarrow \dfrac{\left( b+c \right)}{\left( c+b \right)\left( c+a \right)}-\dfrac{\left( c+a \right)}{\left( c+b \right)\left( c+a \right)}=\dfrac{\left( c+a \right)}{\left( b+a \right)\left( c+a \right)}-\dfrac{\left( a+b \right)}{\left( b+a \right)\left( c+a \right)}
Now, cancelling the common terms from numerator and denominator both sides, we get:
⇒1(c+a)−1(c+b)=1(a+b)−1(c+a)\Rightarrow \dfrac{1}{\left( c+a \right)}-\dfrac{1}{\left( c+b \right)}=\dfrac{1}{\left( a+b \right)}-\dfrac{1}{\left( c+a \right)}
Now let us multiply both sides by (a+b+c) we get:

& \Rightarrow \left( a+b+c \right)\left( \dfrac{1}{c+a}-\dfrac{1}{c+b} \right)=\left( a+b+c \right)\left( \dfrac{1}{a+b}-\dfrac{1}{c+a} \right) \\\ & \Rightarrow \dfrac{a+b+c}{\left( c+a \right)}-\dfrac{a+b+c}{\left( c+b \right)}=\dfrac{a+b+c}{\left( a+b \right)}-\dfrac{a+b+c}{\left( c+a \right)} \\\ & \Rightarrow \dfrac{\left( c+a \right)+b}{\left( c+a \right)}-\dfrac{a+\left( c+b \right)}{\left( c+b \right)}=\dfrac{\left( a+b \right)+c}{\left( a+b \right)}-\dfrac{\left( c+a \right)+b}{\left( c+a \right)} \\\ \end{aligned}$$ Separating the terms again, we get: $$\Rightarrow \dfrac{\left( c+a \right)}{\left( c+a \right)}+\dfrac{b}{\left( c+a \right)}-\dfrac{a}{\left( c+b \right)}-\dfrac{\left( c+b \right)}{\left( c+b \right)}=\dfrac{\left( a+b \right)}{\left( a+b \right)}+\dfrac{c}{\left( a+b \right)}-\dfrac{\left( c+a \right)}{\left( c+a \right)}-\dfrac{b}{\left( c+a \right)}$$ Cancelling out common terms from numerator and denominator, we get: $$\Rightarrow 1+\dfrac{b}{\left( c+a \right)}-\dfrac{a}{\left( c+b \right)}-1=1+\dfrac{c}{\left( a+b \right)}-1-\dfrac{b}{\left( c+a \right)}$$ Simplifying and removing 1 from both sides, we get: $$\Rightarrow \dfrac{b}{c+a}-\dfrac{a}{c+b}=\dfrac{c}{a+b}-\dfrac{b}{c+a}$$ As we can see from equation above, if $\dfrac{a}{c+b}$ was to be the first term, $\dfrac{b}{c+a}$ to be the second term and $\dfrac{c}{a+b}$ to be third term, then it should suggest the difference between second and first term to be equal to difference between third and second term. Therefore, they are in arithmetic progression. Hence, $\dfrac{a}{b+c},\dfrac{b}{c+a},\dfrac{c}{a+b}$ are in arithmetic progression. Hence proved. **Note:** Students should take care while performing every step because the calculations and equation are complex. Students can make mistakes in signs in these questions. While applying ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ make sure that negative sign is with b in both sides. To prove three times to be in AP, it is sufficient to prove that the difference between third term and second term is equal to difference between second term and first term.