Question

Prove that if for an exponential function $ y = {a^x}\left( {a > 0;a \ne 1} \right) $ the values of the argument $ x = {x_n}\left( {n = 1,2,...} \right) $ form an arithmetic progression, then the corresponding values of the function $ {y_n} = {a^{{x_n}}}\left( {n = 1,2,...} \right) $ form a geometric progression.

Answer 1

Hint: When the arithmetic difference between any two successive terms in a series is a constant then the series is said to be in an Arithmetic Progression. For example, a series given by $ 1,2,3,4,5,... $ is an Arithmetic Progression because the difference between its two successive terms is always $ 1 $ .
Also, when the ratio between any two successive terms in a series is a constant then the series is said to be in a Geometric Progression. For example, a series given by $ 2,4,6,8,10,... $ is a Geometric Progression because the ratio between its two successive terms is always 2.

Complete step-by-step answer:
The values of the argument $ x = {x_n}\left( {n = 1,2,...} \right) $ which means its values for $ n = 1,2,3,4,... $ are given by $ {x_1},{x_2},{x_3},{x_4}... $ respectively.
Since these values are in Arithmetic Progression so,
  $ \begin{array}{c}
\left( {{x_2} - {x_1}} \right) = d\\
\left( {{x_3} - {x_2}} \right) = d\\
\left( {{x_4} - {x_3}} \right){\rm{ = d }}
\end{array} $
Where, $ d $ is the difference between two successive terms in series.
Now, we know the given function is,
  $ {y_n} = {a^{{x_n}}}\left( {n = 1,2,...} \right) $
For this function to form a Geometric Progression, the necessary condition is given by,
  $ \dfrac{{{y_2}}}{{{y_1}}} = \dfrac{{{y_3}}}{{{y_2}}} $
And
  $ \dfrac{{{y_3}}}{{{y_2}}} = \dfrac{{{y_4}}}{{{y_3}}} $
We also know that the value of this function is,
  $ {y_n} = {a^{{x_n}}} $ so, for $ n = 1,2,3... $ we have,
  $ \begin{array}{c}
{y_1} = {a^{{x_1}}}\\
{y_2} = {a^{{x_2}}}\\
{y_3} = {a^{{x_3}}}
\end{array} $
Using the condition for Geometric Progression we get,
  $ \dfrac{{{y_2}}}{{{y_1}}} = \dfrac{{{a^{{x_2}}}}}{{{a^{{x_1}}}}} $
We can also write this as,
  $ \dfrac{{{a^{{x_2}}}}}{{{a^{{x_1}}}}} = {a^{{x_2} - {x_1}}} $
And we know that for an Arithmetic Progression,
  $ \left( {{x_2} - {x_1}} \right) = d $
So, substituting this we get,
  $ {a^{{x_2} - {x_1}}} = {a^d} $
Which means that
  $ \dfrac{{{y_2}}}{{{y_1}}} = {a^d} $
Similarly, for $ n = 2,3... $ we have,
  $ {a^{{x_3} - {x_2}}} = {a^d} $ which means that,
  $ \dfrac{{{y_3}}}{{{y_2}}} = {a^d} $
Now it is clear that for $ n = 1,2,3... $ , we have,
  $ \begin{array}{c}
\dfrac{{{y_2}}}{{{y_1}}} = \dfrac{{{y_3}}}{{{y_2}}}\\
{y_1}{y_3} = {y_2}^2
\end{array} $
The above expression says that the terms $ {y_1},{y_2},{y_3},... $ form a Geometric Progression.
Therefore, it is proved that for an exponential function form a geometric progression.

Note: For a series in Arithmetic Progression it is also possible that its exponents are in Geometric Progression or vice-versa. This type of series is also known as the Combination progression or mixed Progression.