Question

Let $b_i>1\text{for}i=1,2,...,101.$ Suppose ${\log }_e{b}_1,{\log }_e{b}_2,...,{\log }_e{b}_{101}$ are in Arithmetic Progression (A.P.) with the common difference ${\log }_{e}2.$ Suppose $a_1,a_2,...,a_{101}$ are in A.P. such that $a_1=b_1$ and $a_{51}=b_{51}.$ If $t=b_1+b_2+\cdots +b_{51}\text{and}s=a_1+a_2+\cdots +a_{51},$ then

• A. $s>t\text{and}{a}_{101}>b_{101}$
• B. $s>t\text{and}{a}_{101}<b_{101}$
• C. $s<t\text{and}{a}_{101}>b_{101}$
• D. $s<t\text{and}{a}_{101}<b_{101}$

Answer 1

The correct option is B $s>t\text{and}{a}_{101}<b_{101}$

As ${\log }_e{b}_1,{\log }_e{b}_2,.....,{\log }_e{b}_{101}$ are in A.P. with common difference ${\log }_{e}2,$
$\text{So}{\log }_e{b}_{i+1}-{\log }_e{b}_i={\log }_{e}2,i=1,2,...,100\Rightarrow b_{i+1}=2b_i$
$\therefore b_1,b_2,...,b_{101}$ are in Geometric Progression(G.P.) with common ratio $2.$

$\text{Let}{a}_1=b_1=y$ $\text{and}{a}_{51}=b_{51}=x=b_1{.2}^{51-1}$
$\text{then}x=y{.2}^{50}$
$s=\frac{51}{2}(a_1+a_{51})=\frac{51(x+y)}{2}=\frac{51(y{.2}^{50}+y)}{2}$

$t=b_1(2^{51}-1)=y(2^{51}-1)$

$\text{Now}s-t={51.2}^{49}y+\frac{53}{2}y-2^{51}.y={47.2}^{49}y+\frac{53}{2}y>0$
$\text{i.e.,}s-t>0\Rightarrow s>t$

Also, $a_1,a_{51},\text{and}{a}_{101}$ are in A.P.
$\therefore 2a_{51}=a_{101}+a_1\Rightarrow a_{101}=2a_{51}-a_1$
$\Rightarrow a_{101}=2x-y$
and $b_1,b_{51},\text{and}{b}_{101}$ are in G.P.
$\therefore b_{101}=\frac{(b_{51}{)}^2}{b_1}=\frac{x^2}{y}$
$b_{101}-a_{101}=\frac{x^2}{y}-2x+y=\frac{(x-y{)}^2}{y}>0$
$\therefore b_{101}>a_{101}$