Mathematics - Sequence & Series Question with Solution | TestHub

If${\log }_e{b}_1,{\log }_e{b}_2\dots .{\log }_e{b}_{101}\to AP; \; \; \mathrm{difference(d)}={\log }_{e}2$
$\Rightarrow b_1,b_2,b_3\mathrm{.......}{b}_{101}\to GP;r=2$
$\therefore b_1,2b_1,2^2{b}_1\mathrm{.........},2^{100}{b}_1\to GP$
$a_1,a_2,a_3\mathrm{..........}{a}_{101}\to AP$Let Common Difference = D
Given,$a_1=b_1$and$a_{51}=b_{51}$
$\Rightarrow$$a_1+50D=2^{50}{b}_1$
$\therefore a_1+50D=2^{50}{a}_1\left(As{b}_1=a_1\right)\Rightarrow D=\frac{2^{50}{a}_1-a_1}{50}$
$t=b_1+b_2+\mathrm{.......}{b}_{51}=b_1+2b_1+2^2{b}_1+{\mathrm{.......2}}^{50}{b}_1\Rightarrow t=b_1\left(2^{51}-1\right);$
$s=a_1+a_2+\mathrm{.......}{a}_{51}\Rightarrow s=\frac{51}{2}\left(2a_1+50D\right)$
$t=a_1. 2^{51}-a_1\Rightarrow t<a_1. 2^{51}$.......(i)$\left(as{b}_i>1 \& a_1=b_{1}so{a}_1>1\right)$
$s=\frac{51}{2} \left(a_1+a_1+50D\right)$
$s=\frac{51}{2} \left(a_1+2^{50} a_1\right)$
$s=\frac{51a_1}{2}+\frac{51}{2}.2^{50} a_1$
$\Rightarrow$$s>a_1. 2^{51}$......(ii)
Clearly$s>t$(from equation (i) and (ii))
Also$a_{101}=a_1+100D; b_{101}=b_1. 2^{100}$
$\therefore$$a_{101}=a_1+100 \left(\frac{2^{50} a_1-a_1}{50}\right); b_{101}=2^{100} a_1$.......(ii)
$a_{101}=a_1+2^{51} a_1-2a_1 \Rightarrow a_{101}=2^{51} a_1-a_1 \Rightarrow a_{101}<2^{51} a_1$.......(iv)
Clearly$b_{101}>a_{101}$(from equation (iii) and (iv))