Question

Let $a_i=i+\frac{1}{t}$ for $i=1,2,......,20$ put $p=\frac{1}{20}(a_1+a_2+.....+a_{20})$ and $q=\frac{1}{20}\left(\frac{1}{a_1}+\frac{1}{a_2}+.....+\frac{1}{a_{20}}\right)$. Then,

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

Answer 1

Answer: (A)

$qϵ(0,\frac{22-p}{21})$

It is evident that $p$ is the AM while $q$ is $\frac{1}{\text{HM}}$ of $a_1,a_2,\dots a_{20}$.

Case I:

Now, consider $t=1$ for simplification.

$\therefore p=\frac{1}{20}(2+3+\cdots +21)=\frac{23}{2}$

$⟹\frac{22-p}{21}=\frac{1}{2}$

Also,

$q=\frac{1}{20}(\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{21})$

Now, replace the fractions by nearest higher valued powers of $\frac{1}{2}$

$⟹q\le \frac{1}{20}(\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{16})$

$⟹q\le \frac{1}{20}(2\times \frac{1}{2}+4\times \frac{1}{4}+8\times \frac{1}{8}+6\times \frac{1}{16})$

$⟹q\le \frac{1}{20}(3+\frac{3}{8})$

$⟹q\le \frac{27}{160}<\frac{22-p}{21}$

Case II:

As the value of $t$ increases beyond 1, the value of $a_i$ decreases and hence, we should also consider a term when $\frac{1}{t}$ is negligible.

$p=\frac{23}{2}$

$⟹\frac{22-p}{21}=\frac{23}{42}$

$q=\frac{1}{20}(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{20})$

$⟹q\le \frac{1}{20}(1+\frac{1}{2}+\frac{1}{2}+\cdots +\frac{1}{16})$

$⟹q\le \frac{1}{20}(4+\frac{5}{16})$

$⟹q\le \frac{69}{320}<\frac{22-p}{21}$

Case III:

As the value of $t$ goes below 1, $\frac{1}{t}$ increases and hence, $a_i$ increases. Therefore $p$ increases and $q$ decreases again satisfying the above equality.

Hence, the only possible answer is $q\in (0,\frac{22-p}{21})$