QJEEAdv16Maa91d: JEE Advanced 2016 Mathematics - Class 11 JEEAdv16Maa91d Solution: Let b i > 1 for I = 1, 2, ......, 101. Suppose log e b 1 , log…

If logb₁, logb₂, ......, logb₁₀₁ are in A.P. with common difference log_e2, then b₁, b₂, ......, b₁₀₁ are in G.P., with common ratio 2.

$\therefore$ b₁ = 2⁰b₁

b₂ = 2¹b₁

b₃ = 2²b₁

$$\begin{array}{ccc}⋮& ⋮& ⋮\end{array}$$

b₁₀₁ = 2¹⁰⁰b₁ ..... (i)

Also, a₁, a₂, ......, a₁₀₁ are in A.P.

Given, a₁ = b₁ and a₅₁ = b₅₁

$\Rightarrow$ a₁ = b₁ and a₅₁ = b₅₁

$\Rightarrow$ a₁ + 50D = 2⁵⁰ b₁

$\Rightarrow$ a₁ + 50D = 2⁵⁰ a₁ [ $\because$ a₁ = b₁] ...... (ii)

Now, t = b₁ + b₂ + ..... + b₅₁

$\Rightarrow t = {b_1}{{({2^{51}} - 1)} \over {2 - 1}}$ ..... (iii)

and s = a₁ + a₂ + .... + a₅₁

$= {{51} \over 2}(2{a_1} + 50D)$ ...... (iv)

$\therefore$ t = a₁(2⁵¹ $-$ 1) [ $\because$ a₁ = b₁]

or t = 2⁵¹ a₁ $-$ a₁ < 2⁵¹ a₁ ...... (v)

and $s = {{51} \over 2}[{a_1} + ({a_1} + 50D)]$ [from Eq. (ii)]

$= {{51} \over 2}[{a_1} + {2^{50}}{a_1}] = {{51} \over 2}{a_1} + {{51} \over 2}{2^{50}}{a_1}$

$\therefore$ s > 2⁵¹ a₁ ...... (vi)

From Eqs. (v) and (vi),

we get s > t

Also, a₁₀₁ = a₁ + 100 D

and b₁₀₁ = 2¹⁰⁰ b₁

$\therefore$ ${a_{101}} = {a_1} + 100\left( {{{{2^{50}}{a_1} - {a_1}} \over {50}}} \right)$

and b₁₀₁ = 2¹⁰⁰ a₁

$\Rightarrow$ a₁₀₁ = a₁ + 2⁵¹ a₁ $-$ 2a₁ = 2⁵¹ a₁ $-$ a₁

$\Rightarrow$ a₁₀₁ < 2⁵¹ a₁

and b₁₀₁ > 2⁵¹ a₁ $\Rightarrow$ b₁₀₁ > a₁₀₁