Question

Let $b_i>1$ 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 $t=b_1+b_2+....+b_{51}$ and $s=a_1+a_2+....+a_{51}$, then

• A. $s>t$ and $a_{101}>b_{101}$
• B. $s>t$ and $a_{101}<b_{101}$
• C. $s<t$ and $a_{101}>b_{101}$
• D. $s<t$ and $a_{101}<b_{101}$

Answer 1

Answer: (B)

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

$lo{g}_e{b}_1,lo{g}_e{b}_2,lo{g}_e{b}_3,......lo{g}_e{b}_{101}$ are in A.P.
$b_1,b_2,b_3,...........,b_{101}$ are in G.P.
Given : $lo{g}_e\left(b_2\right)-lo{g}_e\left(b_1\right)=lo{g}_e\left(2\right)\Rightarrow \frac{b_2}{b_1}=2=r$ (common ratio of G.P. )
$a_1,a_2,a_3,.........a_{101}$ are in A.P.
$a_1=b_1=a$
$b_1+b_2+b_3+........b_{51}=t$ ,
$S=a_1+a_2+......+a_{51}$
$t=$ sum of $51$ terms of G.P. $=b_1\frac{(r^{51}-1)}{r-1}=\frac{a(2^{51}-1)}{2-1}=a(2^{51}-1)$
$s=$ sum of $51$ terms of A.P $=\frac{51}{2}]2a_1+(n-1\left)d\right]=\frac{51}{2}(2a+50d)$
Given $a_{51}=b_{51}$
$a+50d=a(2{)}^{50}$
$50d=a(2^{50}-1)$
Hence, $\Rightarrow s=a({51.2}^{49}+\frac{51}{2})$
$s=2({4.2}^{49}+{47.2}^{49}+\frac{51}{2})\Rightarrow s=a\left(\right(2^{51}-1)+{47.2}^{49}+\frac{53}{2})$
$s-t=a({47.2}^{49}+\frac{53}{2})$
Clearly $s>t$
$a_{101}=a_1+100d=a+2a{.2}^{50}2a=a(2^{51}-1)$
$b_{101}=b_1{r}^{100}=a{.2}^{100}$

Hence : $b_{101}>a_{101}$