Let b_i 1 for i = 1, 2, ldots, 101. Suppose | vedclass

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(B) Given $\log _e b_1, \log _e b_2, \ldots, \log _e b_{101}$ are in $A.P.$,it follows that $b_1, b_2, \ldots, b_{101}$ are in $G.P.$ with common rati

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$s > t$ and $a_{101} < b_{101}$

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(B) Given $\log _e b_1, \log _e b_2, \ldots, \log _e b_{101}$ are in $A.P.$,it follows that $b_1, b_2, \ldots, b_{101}$ are in $G.P.$ with common ratio $r = e^{\log _e 2} = 2$.Let $a_1 = b_1 = a$. Since $a_{51} = b_{51}$,we have $a + 50d = a \cdot r^{50} = a \cdot 2^{50}$,so $d = \frac{a(2^{50} - 1)}{50}$.The sum $t$ of the first $51$ terms of the $G.P.$ is $t = b_1 \frac{r^{51} - 1}{r - 1} = a(2^{51} - 1)$.The sum $s$ of the first $51$ terms of the $A.P.$ is $s = \frac{51}{2}(2a + 50d) = \frac{51}{2}(2a + a(2^{50} - 1)) = \frac{51}{2}a(2^{50} + 1)$.Comparing $s$ and $t$: $s = \frac{51}{2}a(2^{50} + 1) \approx 25.5 \cdot a \cdot 2^{50}$ and $t = a(2^{51} - 1) \approx 2 \cdot a \cdot 2^{50}$. Thus,$s > t$.For the $101^{st}$ terms: $a_{101} = a + 100d = a + 2a(2^{50} - 1) = a(2^{51} - 1)$.$b_{101} = b_1 \cdot r^{100} = a \cdot 2^{100}$.Since $2^{100} > 2^{51} - 1$,we have $b_{101} > a_{101}$.

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