Question

Let $a_i=i+\frac{1}{i}$ for $i=1,2,...,20.$ Put
$p=\frac{1}{20}(a_1+a_2+\cdots +a_{20})\text{and}q=\frac{1}{20}(\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_{20}}).$
Then

• A. $q\in (0,\frac{22-p}{21})$
• B. $q\in (\frac{22-p}{21},\frac{2(22-p)}{21})$
• C. $q\in (\frac{2(22-p)}{21},\frac{22-p}{7})$
• D. $q\in (\frac{22-p}{7},\frac{4(22-p)}{21})$

Answer 1

The correct option is A $q\in (0,\frac{22-p}{21})$

Since $AM\ge GM$
$a_i\ge 2$ for each $i=1,2,...,20$
$\Rightarrow p\ge \frac{1}{20}(2\times 20)$
$\Rightarrow p\ge 2\cdots \left(1\right)$

Since the factor $\frac{22-p}{21}$ is common in all the options, we try to find bound for $\frac{22-p}{21}.$
$\frac{22-p}{21}\le \frac{20}{21}\left[\text{From (1)}\right]$

$\frac{1}{a_i}\le \frac{1}{2}$
$\Rightarrow q\le \frac{1}{20}(20\times \frac{1}{2})$
$\Rightarrow q\le \frac{1}{2}$

$\therefore$ Option (a) is correct.