Question

If \[{{a}^{2}},{{b}^{2}},{{c}^{2}}\] are in A.P. consider two statements
(i) \[\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b}\] are in A.P
(ii) \[\dfrac{a}{b+c},\dfrac{b}{c+a},\dfrac{c}{a+b}\] are in A.P
A. (i) and (ii) both correct
B. (i) and (ii) both incorrect
C. (i) correct (ii) incorrect
D. (i) incorrect (ii) correct

Answer 1

Hint: In this question, we have to prove that the given series are in A.P. Here we have given with a series which is A.P. So, we can find its arithmetic mean. By using this A.M in the given series, we can prove that the given series is also in A.P. Here the arithmetic mean is half of the first and last terms of the series which is equal to the middle term.

Formula Used: If $a,b,c$ are in arithmetic progression, then their arithmetic mean is half of the first and last terms of the series and it is equal to the middle term of the series.
$\begin{align}
  & A.M=\dfrac{a+c}{2}=b \\
 & \Rightarrow a+c=2b \\
\end{align}$

Complete step by step solution: Given that, \[{{a}^{2}},{{b}^{2}},{{c}^{2}}\] are in arithmetic progression.
Then, their arithmetic mean is
$\begin{align}
  & \dfrac{{{a}^{2}}+{{c}^{2}}}{2}={{b}^{2}} \\
 & \Rightarrow {{a}^{2}}+{{c}^{2}}=2{{b}^{2}}\text{ }...(1) \\
\end{align}$
Then, the given series,
If \[\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b}\] are in arithmetic progression, we can write their arithmetic mean as
$\dfrac{1}{b+c}+\dfrac{1}{a+b}=\dfrac{2}{c+a}$
Then, on simplifying the above expression, we get
$\begin{align}
  & \dfrac{1}{b+c}+\dfrac{1}{a+b}=\dfrac{2}{c+a} \\
 & \Rightarrow \dfrac{(a+b)+(b+c)}{(b+c)(a+b)}=\dfrac{2}{c+a} \\
 & \Rightarrow \dfrac{(a+2b+c)}{ab+{{b}^{2}}+ac+bc}=\dfrac{2}{c+a} \\
 & \Rightarrow \dfrac{(a+2b+c)(c+a)}{ab+{{b}^{2}}+ac+bc}=2 \\
\end{align}$
$\begin{align}
  & \Rightarrow \dfrac{ac+{{a}^{2}}+2bc+2ab+{{c}^{2}}+ac}{{{b}^{2}}+ab+ac+bc}=2 \\
 & \Rightarrow \dfrac{{{a}^{2}}+{{c}^{2}}+2ab+2ac+2bc}{{{b}^{2}}+ab+ac+bc}=2 \\
\end{align}$
On substituting (1), we get
$\begin{align}
  & \Rightarrow \dfrac{2{{b}^{2}}+2ab+2ac+2bc}{{{b}^{2}}+ab+ac+bc}=2 \\
 & \Rightarrow \dfrac{2\left( {{b}^{2}}+ab+ac+bc \right)}{{{b}^{2}}+ab+ac+bc}=2 \\
 & \therefore 2=2 \\
\end{align}$
Hence, the given terms \[\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b}\] are in arithmetic progression.
So, statement (i) is correct.
Now, the second statement where, if the terms \[\dfrac{a}{b+c},\dfrac{b}{c+a},\dfrac{c}{a+b}\] are in arithmetic progression, we can write their arithmetic mean as
\[\dfrac{a}{b+c}+\dfrac{c}{a+b}=\dfrac{2b}{c+a}\]
The, on simplifying the above expression, we get
\[\begin{align}
  & \dfrac{a}{b+c}+\dfrac{c}{a+b}=\dfrac{2b}{c+a} \\
 & \Rightarrow \dfrac{a(a+b)+c(b+c)}{(b+c)(a+b)}=\dfrac{2b}{c+a} \\
 & \Rightarrow \dfrac{{{a}^{2}}+ab+cb+{{c}^{2}}}{ab+{{b}^{2}}+ac+bc}=\dfrac{2b}{c+a} \\
 & \Rightarrow \dfrac{({{a}^{2}}+ab+cb+{{c}^{2}})(c+a)}{b(ab+{{b}^{2}}+ac+bc)}=2 \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \dfrac{({{a}^{2}}+{{c}^{2}}+ab+cb)(c+a)}{b(ab+{{b}^{2}}+ac+bc)}=2 \\
 & \Rightarrow \dfrac{\left( 2{{b}^{2}}+ab+bc \right)(c+a)}{b(ab+{{b}^{2}}+ac+bc)}=2 \\
 & \Rightarrow \dfrac{2{{b}^{2}}c+2a{{b}^{2}}+abc+{{a}^{2}}b+b{{c}^{2}}+abc}{b(ab+{{b}^{2}}+ac+bc)}=2 \\
 & \Rightarrow \dfrac{2{{b}^{2}}c+2a{{b}^{2}}+2abc+b({{a}^{2}}+{{c}^{2}})}{b(ab+{{b}^{2}}+ac+bc)}=2 \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \dfrac{2{{b}^{2}}c+2a{{b}^{2}}+2abc+b(2{{b}^{2}})}{b(ab+{{b}^{2}}+ac+bc)}=2 \\
 & \Rightarrow \dfrac{2b(bc+ab+ac+{{b}^{2}})}{b(ab+{{b}^{2}}+ac+bc)}=2 \\
 & \therefore 2=2 \\
\end{align}\]
Hence, the given terms \[\dfrac{a}{b+c},\dfrac{b}{c+a},\dfrac{c}{a+b}\] are in arithmetic progression.
So, statement (ii) is also correct.
Therefore, both statements are correct.

Option ‘A’ is correct

Note: Here we need to remember that, for proving a series to be an arithmetic series, we need to find their arithmetic mean. If the average of the first and last terms of the series is equal to the middle term, then those terms are said to be in arithmetic progression