Sequences and Series 2 Question 7
8. Let $b _i>1$ for $i=1,2, \ldots, 101$. Suppose $\log _e b _1, \log _e b _2$, $\ldots, \log _e b _{101}$ are in AP with the common difference $\log _e 2$ . Suppose $a _1, a _2, \ldots, a _{101}$ are in AP, such that $a _1=b _1$ and $a _{51}=b _{51}$. If $t=b _1+b _2+\ldots+b _{51} \quad$ and $s=a _1+a _2+\ldots+a _{51}$, then
(2016 Adv.)
(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 $a _{101}<b _{101}$
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Answer:
Correct Answer: 8. (c)
Solution:
- If $\log b _1, \log b _2, \ldots, \log b _{101}$ are in $AP$, with common difference $\log _e 2$, then $b _1, b _2, \ldots, b _{101}$ are in GP, with common ratio 2 .
$\therefore b _1=2^{0} b _1, b _2=2^{1} b _1, b _3=2^{2} b _1, \ldots, b _{101}=2^{100} b _1$
Also, $a _1, a _2, \ldots, a _{101}$ are in AP.
Given, $\quad a _1=b _1$ and $a _{51}=b _{51}$
$$ \begin{array}{llll} \Rightarrow & a _1+50 D=2^{50} b _1 \\ \Rightarrow & a _1+50 D=2^{50} a _1 \quad\left[\because a _1=b _1\right] . \end{array} $$
Now, $\quad t=b _1+b _2+\ldots+b _{51}$
$\Rightarrow \quad t=b _1 \frac{\left(2^{51}-1\right)}{2-1}$
and $\quad s=a _1+a _2+\ldots+a _{51}$
$$ =\frac{51}{2}\left(2 a _1+50 D\right) $$
$$ \therefore \quad t=a _1\left(2^{51}-1\right) \quad\left[\because a _1=b _1\right] $$
or $\quad t=2^{51} a _1-a _1<2^{51} a _1$
and $s=\frac{51}{2}\left[a _1+\left(a _1+50 D\right)\right] \quad$ [from Eq. (ii)]
$$ =\frac{51}{2}\left[a _1+2^{50} a _1\right] $$
$$ =\frac{51}{2} a _1+\frac{51}{2} 2^{50} a _1 $$
$\therefore \quad s>2^{51} a _1$
From Eqs. (v) and (vi), we get $s>t$
Also, $\quad a _{101}=a _1+100 D$ and $b _{101}=2^{100} b _1$
$$ \begin{aligned} & \therefore \quad a _{101}=a _1+100 \frac{2^{50} a _1-a _1}{50} \text { and } b _{101}=2^{100} a _1 \\ & \Rightarrow \quad a _{101}=a _1+2^{51} a _1-2 a _1=2^{51} a _1-a _1 \\ & \Rightarrow \quad a _{101}<2^{51} a _1 \text { and } b _{101}>2^{51} a _1 \\ & \Rightarrow \quad b _{101}>a _{101} \end{aligned} $$
