Question

Let a,b be distinct positive real numbers, whose geometric mean equals
$( {a}^{t - 99} + {b}^{t - 99} ) \div ( {a}^{t - 100} + {b}^{t - 100})$
Then t must be equal to:

a) 199/2
b)199
c)99/2
d)99​

Answer 1

We're given two distinct positive integers $a$ and $b$ such that,

$\longrightarrow\sqrt{ab}=\dfrac{a^{t-99}+b^{t-99}}{a^{t-100}+b^{t-100}}\quad\quad\dots(1)$

Since $a^{m-n}=\dfrac{a^m}{a^n},$

$\longrightarrow\sqrt{ab}=\dfrac{\left(\dfrac{a^t}{a^{99}}+\dfrac{b^t}{b^{99}}\right)}{\left(\dfrac{a^t}{a^{100}}+\dfrac{b^t}{b^{100}}\right)}$

$\longrightarrow\sqrt{ab}=\dfrac{\left(\dfrac{a^tb^{99}+a^{99}b^t}{a^{99}b^{99}}\right)}{\left(\dfrac{a^tb^{100}+a^{100}b^t}{a^{100}b^{100}}\right)}$

$\longrightarrow\sqrt{ab}=\dfrac{a^tb^{99}+a^{99}b^t}{a^{99}b^{99}}\cdot\dfrac{a^{100}b^{100}}{a^tb^{100}+a^{100}b^t}$

$\longrightarrow\sqrt{ab}=\dfrac{a^tb^{99}+a^{99}b^t}{(ab)^{99}}\cdot\dfrac{(ab)^{100}}{a^tb^{100}+a^{100}b^t}$

$\longrightarrow\sqrt{ab}=\dfrac{a^tb^{99}+a^{99}b^t}{a^tb^{100}+a^{100}b^t}\cdot ab$

$\longrightarrow\dfrac{1}{\sqrt{ab}}=\dfrac{a^tb^{99}+a^{99}b^t}{a^tb^{100}+a^{100}b^t}$

$\longrightarrow\sqrt{ab}=\dfrac{a^tb^{100}+a^{100}b^t}{a^tb^{99}+a^{99}b^t}$

$\longrightarrow\sqrt{ab}=\dfrac{a^tb^t(b^{100-t}+a^{100-t})}{a^tb^t(b^{99-t}+a^{99-t})}$

$\longrightarrow\sqrt{ab}=\dfrac{a^{100-t}+b^{100-t}}{a^{99-t}+b^{99-t}}$

From (1),

$\longrightarrow\dfrac{a^{t-99}+b^{t-99}}{a^{t-100}+b^{t-100}}=\dfrac{a^{100-t}+b^{100-t}}{a^{99-t}+b^{99-t}}$

$\longrightarrow(a^{t-99}+b^{t-99})(a^{99-t}+b^{99-t})=(a^{t-100}+b^{t-100})(a^{100-t}+b^{100-t})$

$\longrightarrow a^{t-99}b^{99-t}+a^{99-t}b^{t-99}=a^{t-100}b^{100-t}+a^{100-t}a^{t-100}$

$\longrightarrow a^{t-99}(b^{-1})^{t-99}+a^{99-t}(b^{-1})^{99-t}=a^{t-100}(b^{-1})^{t-100}+a^{100-t}(b^{-1})^{100-t}$

$\longrightarrow \left(\dfrac{a}{b}\right)^{t-99}+\left(\dfrac{a}{b}\right)^{99-t}=\left(\dfrac{a}{b}\right)^{t-100}+\left(\dfrac{a}{b}\right)^{100-t}$

$\longrightarrow \left(\dfrac{a}{b}\right)^{t-99}+\left(\dfrac{a}{b}\right)^{99-t}=\left(\dfrac{a}{b}\right)^{t-99-1}+\left(\dfrac{a}{b}\right)^{1+99-t}$

Taking $\dfrac{a}{b}=x$ and $t-99=n,$

$\longrightarrow x^n+x^{-n}=x^{n-1}+x^{1-n}$

$\longrightarrow x^n-x^{n-1}=x^{1-n}-x^{-n}$

$\longrightarrow x^n(1-x^{-1})=x^{-n}(x-1)$

$\longrightarrow x^n\left(1-\dfrac{1}{x}\right)=x^{-n}(x-1)$

$\longrightarrow x^n\left(\dfrac{x-1}{x}\right)=x^{-n}(x-1)$

$\longrightarrow x^{n-1}=x^{-n}$

Equating powers since bases are same,

$\longrightarrow n-1=-n$

$\longrightarrow n=\dfrac{1}{2}$

$\longrightarrow t-99=\dfrac{1}{2}$

$\longrightarrow t=99+\dfrac{1}{2}$

$\longrightarrow \underline{\underline{t=\dfrac{199}{2}}}$

Hence (a) is the answer.

Answer 2

Given : a,b be distinct positive real numbers, whose geometric mean equals $\dfrac{a^{t-99} + b^{t-99}}{a^{t-100} + b^{t-100}}$  

To Find : Value of t

Solution:

$\dfrac{a^{t-99} + b^{t-99}}{a^{t-100} + b^{t-100}}$

is geometric mean of ab

hence :

$\sqrt{ab} = \dfrac{a^{t-99} + b^{t-99}}{a^{t-100} + b^{t-100}}$

let say t-100 = n

=> t - 99 = n + 1

=> √ab  = (aⁿ⁺¹ + bⁿ⁺¹ ) / (aⁿ+ bⁿ )

=> √ab (aⁿ+ bⁿ ) =  (aⁿ⁺¹ + bⁿ⁺¹ )

Squaring both sides

=> ab ( a²ⁿ + b²ⁿ + 2aⁿbⁿ) = a²ⁿ⁺² + b²ⁿ⁺² + 2aⁿ⁺¹bⁿ⁺¹

=> ba²ⁿ⁺¹ + ab²ⁿ⁺¹ + 2aⁿ⁺¹bⁿ⁺¹ = a²ⁿ⁺² + b²ⁿ⁺² + 2aⁿ⁺¹bⁿ⁺¹

=> ba²ⁿ⁺¹ + ab²ⁿ⁺¹  = a²ⁿ⁺² + b²ⁿ⁺²

=> a²ⁿ⁺¹ (b - a)  + b²ⁿ⁺¹(a-b) = 0

=>  a²ⁿ⁺¹ (b - a)  - b²ⁿ⁺¹(b-a) = 0

=> (a²ⁿ⁺¹ -  b²ⁿ⁺¹) (b - a) = 0

=> a²ⁿ⁺¹ =  b²ⁿ⁺¹   as a & b are distinct numbers so b - a ≠ 0

=>2n+1 = 0  as a & b are distinct numbers

=> n = -1/2

t-100 = n

=> t = n + 100

=> t = -1/2 + 100

=> t = 199/2

t must be equal to  199/2

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