Question

If $ x = \sum\limits_{n = 0}^n {{a^n}} $ , $ y = \sum\limits_{n = 0}^n {{b^n}} $ , $ z = \sum\limits_{n = 0}^n {{c^n}} $ where $ a $ , $ b $ and $ c $ are in AP and $ \left| a \right| < 1 $ , $ \left| b \right| < 1 $ and $ \left| c \right| < 1 $ then $ x $ , $ y $ , $ z $ are in
A.HP
B.Arithmetic-Geometric progression
C.AP
D.GP

Answer 1

Hint: Since, $ x $ is in geometric mean the formula to find the infinite geometric mean is $ \dfrac{a}{{1 - r}} $ . Where, $ a $ is the first term of the series and $ r $ is the common multiple for the series. Find the geometric mean for all the values for $ x $ , $ y $ , $ z $ . Then find the relation for the $ x $ , $ y $ and $ z $ .

Complete step-by-step answer:
The value of $ x $ is equal to $ \sum\limits_{n = 0}^\infty {{a^n}} $ .
The value of $ y $ is equal to $ \sum\limits_{n = 0}^\infty {{b^n}} $ .
The value of $ z $ is equal to $ \sum\limits_{n = 0}^\infty {{c^n}} $ .
Since, we know that $ x = \sum\limits_{n = 0}^\infty {{a^n}} $ .
So, we can see that $ x $ is in the geometric mean so the formula to find the geometric mean is,
 $ \sum\limits_{n = 0}^\infty {{a^n}} = \dfrac{a}{{1 - r}} $
Where, $ a $ is the first term of the series and $ r $ is the common multiple for the series.
So, for $ \sum\limits_{n = 0}^\infty {{a^n}} $ , $ a $ is 1 and $ r $ is $ a $ .
On substituting the values, we get,
 $ \sum\limits_{n = 0}^\infty {{a^n}} = \dfrac{1}{{1 - a}} $
We know that the value of x is,
 $ \begin{array}{c}
x = \sum\limits_{n = 0}^\infty {{a^n}} \\
 = \dfrac{1}{{1 - a}}
\end{array} $ .......(1)
In the case of $ y $ . We can see that $ y $ is in the geometric mean so the formula to find the geometric mean is,
 $ \sum\limits_{n = 1}^\infty {{a^n}} = \dfrac{a}{{1 - r}} $

Where, $ a $ is the first term of the series and $ r $ is the common multiple for the series.
So for $ \sum\limits_{n = 0}^\infty {{b^n}} $ , $ a $ is 1 and $ r $ is $ a $ .
On substituting the values, we get,
 $ \sum\limits_{n = 0}^\infty {{b^n}} = \dfrac{1}{{1 - b}} $
We know that the value of y is,
 $ \begin{array}{c}
y = \sum\limits_{n = 1}^\infty {{b^n}} \\
 = \dfrac{1}{{1 - b}}
\end{array} $ ......(2)
In the case of $ z $ . So, we can see that $ z $ is in the geometric mean so the formula to find the geometric mean is,
 $ \sum\limits_{n = 1}^\infty {{a^n}} = \dfrac{a}{{1 - r}} $
Where $ a $ is the first term of the series and $ r $ is the common multiple for the series.
So, for $ \sum\limits_{n = 0}^\infty {{c^n}} $ , $ a $ is 1 and $ r $ is $ a $ .
On substituting the values, we get,
 $ \sum\limits_{n = 0}^\infty {{c^n}} = \dfrac{1}{{1 - c}} $
We know that the value of z is,
 $ \begin{array}{c}
z = \sum\limits_{n = 0}^\infty {{c^n}} \\
 = \dfrac{1}{{1 - c}}
\end{array} $ .........(3)
We can rewrite the equation (1) as,
 $ \begin{array}{l}
x = \dfrac{1}{{1 - a}}\\
\dfrac{1}{x} = 1 - a\\
a = 1 - \dfrac{1}{x}
\end{array} $
Hence, we get $ a $ is $ \dfrac{{x - 1}}{x} $ .
We can rewrite the equation (2) as,
 $ \begin{array}{l}
y = \dfrac{1}{{1 - b}}\\
\dfrac{1}{y} = 1 - b\\
b = 1 - \dfrac{1}{y}
\end{array} $
Hence, we get $ b $ is $ \dfrac{{y - 1}}{y} $ .
We can rewrite the equation (3) as,
 $ \begin{array}{l}
z = \dfrac{1}{{1 - c}}\\
\dfrac{1}{z} = 1 - c\\
c = 1 - \dfrac{1}{z}
\end{array} $
Hence, we get $ c $ is $ \dfrac{{z - 1}}{z} $ .
Since we know that $ a $ , $ b $ , $ c $ are in AP.
The formula for $ a $ , $ b $ and $ c $ will be,
 $ ab = a + c $
On substituting the values of $ a $ , $ b $ and $ c $ we get,
 $ \begin{array}{c}
2\left( {\dfrac{{y - 1}}{y}} \right) = \left( {\dfrac{{x - 1}}{x}} \right) + \left( {\dfrac{{z - 1}}{z}} \right)\\
2 - \dfrac{2}{y} = 1 - \dfrac{1}{x} + 1 - \dfrac{1}{y}
\end{array} $
By rearranging the terms, we get that,
 $ \begin{array}{c}
 - \dfrac{2}{y} = - \dfrac{1}{x} - \dfrac{1}{z}\\
\dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z}
\end{array} $
Hence $ \dfrac{1}{x} $ , $ \dfrac{1}{y} $ and $ \dfrac{1}{z} $ are in AP.
This implies that $ x $ , $ y $ and $ z $ are in HP.
So, the correct answer is “Option A”.

Note: If the sequence terms are like $ \dfrac{1}{x} $ , $ \dfrac{1}{y} $ , $ \dfrac{1}{z} $ are in arithmetic mean then $ x $ , $ y $ and $ z $ will be in harmonic mean. The relation between arithmetic mean, harmonic mean and geometric mean is given by $ \left( {{\rm{AM}}} \right) \times \left( {{\rm{HM}}} \right) = {\left( {{\rm{GM}}} \right)^2} $ .