Question

(i) If \[a\left( \dfrac{1}{b}+\dfrac{1}{c} \right),b\left( \dfrac{1}{c}+\dfrac{1}{a} \right),c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)\] are in A.P. Prove that a, b, c are in A.P.
(ii) If a, b, c are in G.P. Prove that \[\left( {{a}^{n}}+{{b}^{n}} \right),\left( {{b}^{n}}+{{c}^{n}} \right),\left( {{c}^{n}}+{{d}^{n}} \right)\] are in G.P.

Answer 1

Hint: For part (i) of the question, add 1 in each term of the given A.P terms and then simplify the numerator. Cancel the common terms from all the three terms in A.P and multiply the simplified form with abc to get the answer.
For part (ii), assume the four terms as \[a,a{{r}^{1}},a{{r}^{2}},a{{r}^{3}}\], where ‘a’ is the first term and ‘r’ is the common difference of G.P. Simplify the terms \[\left( {{a}^{n}}+{{b}^{n}} \right),\left( {{b}^{n}}+{{c}^{n}} \right),\left( {{c}^{n}}+{{d}^{n}} \right)\] and take the ratio of two terms at a time. If the ratio of \[{{2}^{nd}}\] to \[{{1}^{st}}\] and \[{{3}^{rd}}\] to \[{{2}^{nd}}\] term are equal, then it is proved that the terms are in G.P.

Complete step by step answer:
Let us consider part (i) of the question.
\[\because \]\[a\left( \dfrac{1}{b}+\dfrac{1}{c} \right),b\left( \dfrac{1}{c}+\dfrac{1}{a} \right),c\left( \dfrac{1}{a}+\dfrac{1}{b} \right)\] are in A.P.
Adding 1 to each term, we get,
\[\Rightarrow a\left( \dfrac{1}{b}+\dfrac{1}{c}+1 \right),b\left( \dfrac{1}{c}+\dfrac{1}{a}+1 \right),c\left( \dfrac{1}{a}+\dfrac{1}{b}+1 \right)\] are in A.P.
\[\Rightarrow a\left( \dfrac{b+c+bc}{bc} \right),b\left( \dfrac{a+c+ac}{ac} \right),c\left( \dfrac{a+b+ab}{ab} \right)\] are in A.P.
\[\Rightarrow \left( \dfrac{ab+ac+abc}{bc} \right),\left( \dfrac{ab+bc+abc}{ac} \right),\left( \dfrac{ac+bc+abc}{ab} \right)\] are in A.P.
Cancelling the numerator from each term, we get,
\[\Rightarrow \]\[\left( \dfrac{1}{bc} \right),\left( \dfrac{1}{ac} \right),\left( \dfrac{1}{ab} \right)\] are in A.P.
Multiplying each term with abc, we get,
\[\Rightarrow \]\[\left( \dfrac{abc}{bc} \right),\left( \dfrac{abc}{ac} \right),\left( \dfrac{abc}{ab} \right)\] are in A.P.
\[\Rightarrow \] a, b, c are in A.P.
Now Let us consider part (ii) of the question.
It is given that a, b, c, d are in G.P. Let us consider a, b, c, d as \[a,a{{r}^{1}},a{{r}^{2}},a{{r}^{3}}\] respectively, where ‘a’ is the first term and ‘r’ is the common ratio. Therefore,
\[\begin{align}
  & \Rightarrow a=a \\
 & \Rightarrow b=ar \\
 & \Rightarrow c=a{{r}^{2}} \\
 & \Rightarrow d=a{{r}^{3}} \\
\end{align}\]
Now, \[{{a}^{n}}+{{b}^{n}}={{a}^{n}}+{{a}^{n}}{{r}^{n}}={{a}^{n}}\left[ 1+{{r}^{n}} \right]\]
\[\begin{align}
  & \Rightarrow {{b}^{n}}+{{c}^{n}}={{a}^{n}}{{r}^{n}}+{{a}^{n}}{{r}^{2n}}={{a}^{n}}{{r}^{n}}\left[ 1+{{r}^{n}} \right] \\
 & \Rightarrow {{c}^{n}}+{{d}^{n}}={{a}^{n}}{{r}^{2n}}+{{a}^{n}}{{r}^{3n}}={{a}^{n}}{{r}^{2n}}\left[ 1+{{r}^{n}} \right] \\
\end{align}\]
Considering the ratio of \[{{2}^{nd}}\] term to \[{{1}^{st}}\] term, we get,
\[\Rightarrow \dfrac{{{b}^{n}}+{{c}^{n}}}{{{a}^{n}}+{{b}^{n}}}=\dfrac{{{a}^{n}}{{r}^{n}}\left[ 1+{{r}^{n}} \right]}{{{a}^{n}}\left[ 1+{{r}^{n}} \right]}={{r}^{n}}\]
Considering the ratio of \[{{3}^{rd}}\] term to \[{{2}^{nd}}\] term we get,
\[\Rightarrow \dfrac{{{c}^{n}}+{{d}^{n}}}{{{b}^{n}}+{{c}^{n}}}=\dfrac{{{a}^{n}}{{r}^{2n}}\left[ 1+{{r}^{n}} \right]}{{{a}^{n}}{{r}^{n}}\left[ 1+{{r}^{n}} \right]}={{r}^{n}}\]
Clearly we can see that the ratios are equal. Therefore, the required terms are in G.P.

Note: One may note that we can also solve this question by applying the formula of arithmetic mean in part (i) and formula of geometric mean in part (ii). But then we will have to encounter some difficult calculations. So, it is better to use the above approach.