Question

For what value of n, the geometric mean of a and b is $\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}$.

Answer 1

Hint: In this question use the concept that if we have two numbers a and b then the geometric mean of them will be $\sqrt {ab}$, compare the given geometric mean with this standard mean to get the value of n.

Complete step-by-step answer:
Given geometric mean (G.M) of (a and b) is $\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}$.............. (1)
Now as we know that the G.M of (a) and (b) is $\sqrt {ab}$ ...................... (2)
Therefore both equations (1) and (2) should be equal so equate them we have.
$\Rightarrow \dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \sqrt {ab}$
Now simplify the above equation we have,
$\Rightarrow {a^{n + 1}} + {b^{n + 1}} = {a^{\dfrac{1}{2}}}{b^{\dfrac{1}{2}}}\left( {{a^n} + {b^n}} \right)$
Now again simplify we have,
$\Rightarrow {a^{n + 1}} + {b^{n + 1}} = {a^{\dfrac{1}{2}}}{b^{\dfrac{1}{2}}}{a^n} + {a^{\dfrac{1}{2}}}{b^{\dfrac{1}{2}}}{b^n}$
$\Rightarrow {a^{n + 1}} + {b^{n + 1}} = {a^{n + \dfrac{1}{2}}}{b^{\dfrac{1}{2}}} + {a^{\dfrac{1}{2}}}{b^{n + \dfrac{1}{2}}}$
Now shifting the variables we have,
$\Rightarrow {a^{n + 1}} - {a^{n + \dfrac{1}{2}}}{b^{\dfrac{1}{2}}} = {a^{\dfrac{1}{2}}}{b^{n + \dfrac{1}{2}}} - {b^{n + 1}}$
Now take ${a^{n + \dfrac{1}{2}}}$ common from L.H.S terms and ${b^{n + \dfrac{1}{2}}}$ common from R.H.S terms we have,
$\Rightarrow {a^{n + \dfrac{1}{2}}}\left( {\dfrac{{{a^{n + 1}}}}{{{a^{n + \dfrac{1}{2}}}}} - {b^{\dfrac{1}{2}}}} \right) = {b^{n + \dfrac{1}{2}}}\left( {{a^{\dfrac{1}{2}}} - \dfrac{{{b^{n + 1}}}}{{{b^{n + \dfrac{1}{2}}}}}} \right)$
Now simplify the above equation we have,
$\Rightarrow {a^{n + \dfrac{1}{2}}}\left( {{a^{n + 1 - n - \dfrac{1}{2}}} - {b^{\dfrac{1}{2}}}} \right) = {b^{n + \dfrac{1}{2}}}\left( {{a^{\dfrac{1}{2}}} - {b^{n + 1 - n - \dfrac{1}{2}}}} \right)$
$\Rightarrow {a^{n + \dfrac{1}{2}}}\left( {{a^{\dfrac{1}{2}}} - {b^{\dfrac{1}{2}}}} \right) = {b^{n + \dfrac{1}{2}}}\left( {{a^{\dfrac{1}{2}}} - {b^{\dfrac{1}{2}}}} \right)$
Now cancel out the common terms from L.H.S and R.H.S we have,
$\Rightarrow {a^{n + \dfrac{1}{2}}} = {b^{n + \dfrac{1}{2}}}$
$\Rightarrow \dfrac{{{a^{n + \dfrac{1}{2}}}}}{{{b^{n + \dfrac{1}{2}}}}} = 1$
Now as we know 1 can be written as (a/b)0 so use this property in above equation we have,
$\Rightarrow \dfrac{{{a^{n + \dfrac{1}{2}}}}}{{{b^{n + \dfrac{1}{2}}}}} = {\left( {\dfrac{a}{b}} \right)^0}$
$\Rightarrow {\left( {\dfrac{a}{b}} \right)^{^{n + \dfrac{1}{2}}}} = {\left( {\dfrac{a}{b}} \right)^0}$
So on comparing we have,
$\Rightarrow n + \dfrac{1}{2} = 0$
$\Rightarrow n = - \dfrac{1}{2}$
So $- \dfrac{1}{2}$ is the required value of (n) for which the G.M of (a) and (b) is$\dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}$.
So this is the required answer.

Note: In general there are two means which are most frequently used that is arithmetic mean and the geometric means. A.M between two numbers a and b is $\dfrac{{a + b}}{2}$, the formula of geometric mean is explained above but we can generalize it to n terms as well, the G.M of ${x_1},{x_2},{x_3},{x_4}..........{x_n}{ = ^n}\sqrt {{x_1},{x_2},{x_3},{x_4}..........{x_n}}$.