Question

Let $b_i>1$ for i = 1, 2, .... ,101. Suppose $lo{g}_e{b}_1,lo{g}_e{b}_2,\dots \dots lo{g}_e{b}_{101}$ are in AP with the common difference $lo{g}_{e}2$. Suppose $a_1,a_2,\dots \dots a_{101}$ are in AP, such that $a_1=b_1$ and $a_{51}=b_{51}$. If $t=b_1+b_2+\dots \dots +b_{51}$ and $s=a_1+a_2+\dots \dots +a_{51}$, 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

The correct option is B $s>t$ and $a_{101}<b_{101}$

If $log{b}_1,log{b}_2,\dots \dots log{b}_{101}$ are in AP, with common difference $lo{g}_{e}2$, then $b_1,b_2,\dots \dots b_{101}$ are in GP, with common ratio 2.
$\therefore b_1=2^0{b}_1,b_2=2^1{b}_1,b_3=2^2{b}_1\dots \dots ,b_{101}=2^{100}{b}_1\dots \dots \left(i\right)$
Also, $a_1,a_2\dots \dots a_{101}$ are in AP
Given $a_1=b_1$ and $a_{51}=b_{51}$
$\Rightarrow a_1+50D=2^{50}{b}_1$
$\Rightarrow a_1+50D=2^{50}{a}_1$ $[\because a_1=b_1]\dots \dots \left(ii\right)$
Now, $t=b_1+b_2+\dots \dots +b_{51}$
$\Rightarrow t=b_1\frac{(2^{51}-1)}{2-1}\dots \dots \left(iii\right)$
and $s=a_1+a_2+\dots \dots +a_{51}$
$=\frac{51}{2}(2a_1+50D)\dots \dots \left(iv\right)$
$\therefore t=a_1(2^{51}-1)$ $[\because a_1=b_1]$
or $t=2^{51}{a}_1-a_1<2^{51}{a}_1\dots \dots \left(v\right)$
and $s=\frac{51}{2}[a_1+(a_1+50D\left)\right][fromEq.(ii\left)\right]$
$=\frac{51}{2}[a_1+2^{50}{a}_1]=\frac{51}{2}{a}_1+\frac{51}{2}{2}^{50}{a}_1\therefore s>2^{51}{a}_1\dots \dots \left(vi\right)$
From Eqs. (v) and (vi), we get s > t
Also, $a_{101}=a_1+100Dand{b}_{101}=2^{100}{b}_1$
$\therefore a_{101}=a_1+100\left(\frac{2^{50}{a}_1-a_1}{50}\right)$ and $b_{101}=2^{100}{a}_1$
$\Rightarrow a_{101}=a_1+2^{51}{a}_1-2a_1=2^{51}{a}_1-a_1$
$\Rightarrow a_{101}<2^{51}{a}_{1}and{b}_{101}>2^{51}{a}_1$
$\Rightarrow b_{101}>a_{101}$