Question

If a function given as \[f\left( x \right)={{2}^{x}}\], then \[f\left( 0 \right),f\left( 1 \right),f\left( 2 \right),.....\] are in?
A. AP
B. GP
C. HP
D. Arbitrary

Answer 1

Hint: We will first start by using the fact that if \[f\left( x \right)={{2}^{x}}\] then \[f\left( 0 \right)={{2}^{0}}\]. Then we will similarly find the value of the first few terms of the series. After this we will test the series for AP, GP and HP by finding common differences and common ratios.
Complete step-by-step answer:
Now, we have been given that \[f\left( x \right)={{2}^{x}}\]. Then we have to find the series in which terms \[f\left( 0 \right),f\left( 1 \right),f\left( 2 \right),.....\] are in.
Now, we have,
\[\begin{align}
  & f\left( 0 \right)={{2}^{0}} \\
 & f\left( 1 \right)={{2}^{1}} \\
 & f\left( 2 \right)={{2}^{2}} \\
\end{align}\]
Now, we have the difference of first two terms,
$\begin{align}
  & =f\left( 1 \right)-f\left( 0 \right) \\
 & ={{2}^{1}}-{{2}^{0}} \\
 & =1 \\
\end{align}$
The difference of next two terms,
$\begin{align}
  & =f\left( 2 \right)-f\left( 1 \right) \\
 & ={{2}^{2}}-{{2}^{1}} \\
 & =4-2 \\
 & =2 \\
\end{align}$
Now, we know that if ${{a}_{1}},{{a}_{2}},{{a}_{3}}$ are in AP then their common difference must be same i.e. ${{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}$ but since we have,
$f\left( 1 \right)-f\left( 0 \right)\ne f\left( 2 \right)-f\left( 1 \right)$
Hence, they are not in AP.
Now, we have the common ratio r of the first two terms as $\dfrac{f\left( 1 \right)}{f\left( 0 \right)}=\dfrac{{{2}^{1}}}{{{2}^{0}}}$.
Also, we have the ratio of next two terms as $\dfrac{f\left( 2 \right)}{f\left( 0 \right)}=\dfrac{{{2}^{2}}}{{{2}^{1}}}=2$.
Now, we know that if ${{a}_{1}},{{a}_{2}},{{a}_{3}}$ are in GP. Then their common ratio is same that is $\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}$.
Since we have,
$\dfrac{f\left( 1 \right)}{f\left( 0 \right)}=\dfrac{f\left( 2 \right)}{f\left( 0 \right)}$
There we have the term \[f\left( 0 \right),f\left( 1 \right),f\left( 2 \right),.....\] in GP.
Hence, the correct option is (B).

Note: It is important to note that all the terms in series \[f\left( 0 \right),f\left( 1 \right),f\left( 2 \right),.....f\left( n \right)\] have the common ratio as $\dfrac{f\left( n \right)}{f\left( n-1 \right)}=\dfrac{{{2}^{n}}}{{{2}^{n-1}}}=2$ which is constant in spite of the value of n. Therefore, we have the series as GP. Also, since we have found the series as GP we have not further tested it for HP.