Let b_i1 for i=1,2, ldots, 101. Suppose l | vedclass

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Question

$\log _e b_1, \log _e b_2, \log _e b_3, \ldots \ldots . \log _e b_{101} 	ext { are in A.P. }$

$b_1, b_2, b_3, \ldots \ldots \ldots \ldots, b_{101}

Vedclass · Accepted Answer

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

Vedclass · Answer

$\log _e b_1, \log _e b_2, \log _e b_3, \ldots \ldots . \log _e b_{101} 	ext { are in A.P. }$

$b_1, b_2, b_3, \ldots \ldots \ldots \ldots, b_{101} 	ext { are in G.P. }$

$	ext { Given }: \log _e\left(b_2ight)-\log _e\left(b_1ight)=\log _e(2) \Rightarrow \frac{b_2}{b_1}=2=r 	ext { (common ratio of G.P.) }$

$a_1, a_2, a_3, \ldots \ldots \ldots a_{101} 	ext { are in A.P. }$

$a_1=b_1=a$

$b_1+b_2+b_3+\ldots \ldots \ldots b_{51}=t,$

$S=a_1+a_2+\ldots \ldots+a_{51}$

$t=	ext { sum of } 51 	ext { terms of G.P. }=b_1 \frac{\left(r^{51}-1ight)}{r-1}=\frac{a\left(2^{51}-1ight)}{2-1}=a\left(2^{51}-1ight)$

$\left.\left.s=	ext { sum of } 51 	ext { terms of A.P }=\frac{51}{2}ight] 2 a_1+(n-1) dight]=\frac{51}{2}(2 a+50 d)$

$	ext { Given } a_{51}=b_{51}$

$a+50 d=a(2)^{50}$

$50 d=a\left(2^{50}-1ight)$

$	ext { Hence, } \Rightarrow s=a\left(51.2^{49}+\frac{51}{2}ight)$

$s=2\left(4.2^{49}+47.2^{49}+\frac{51}{2}ight) \Rightarrow$ $s=a\left(\left(2^{51}-1ight)+47.2^{49}+\frac{53}{2}ight)$

$s-t=a\left(47.2^{49}+\frac{53}{2}ight)$

$	ext { Clearly } s > t$

$a_{101}=a_1+100 d=a+2 a .2^{50} 2 a=a\left(2^{51}-1ight)$

$b_{101}=b_1 r^{100}=a \cdot 2^{100}$

Hence $: b_{101} > a_{101}$

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