Question

If $P$ be the $r^{\text{th }}$ term when $n$ arithmetic means are inserted between 5 and 100 and $Q$ be the $r^{\text{th }}$ term when $n$ harmonic means are inserted between 5 and 100 , then find the value of $\left(\frac{P}{5}+\frac{100}{Q}\right)$.

Answer 1

Answer: 0021

Let $x=5$ and $y=100$.
Now, $x,A_1,A_2,A_3,\dots \dots ...,A_n$, $y$ are in A.P. ....(1)
and $x,H_1,H_2,H_2,H_3,\dots ,H_n,y$ are in H.P. ....(2)
Given that, $P=A_{r-1}$ and $Q=H_{r-1}$
From (1), $y=x+(n+1)d\Rightarrow d=\frac{y-x}{n+1}$
$\therefore P=A_{r-1}=r^{\text{th }}\text{ term }=x+(r-1)d=x+(r-1)\cdot \left(\frac{y-x}{n+1}\right)$
$\Rightarrow \frac{P}{x}=1+\frac{(r-1)(y-x)}{x(n+1)}$....(3)
Now, from (2),
$\frac{1}{x},\frac{1}{H_1},\frac{1}{H_2},\dots \dots \dots ,\frac{1}{H_n},\frac{1}{y}$ are in A.P.
$\therefore \frac{1}{y}=(n+2{)}^{\text{th }}\mathrm{term}=\frac{1}{x}+(n+1)d^{\prime }\Rightarrow d^{\prime }=\left(\frac{\frac{1}{y}-\frac{1}{x}}{n+1}\right)$
$\therefore \frac{1}{Q}=\frac{1}{H_{r-1}}=r^{\text{th }}\mathrm{term}=\frac{1}{x}+(r-1)d^{\prime }=\frac{1}{x}+(r-1)\left(\frac{\frac{1}{y}-\frac{1}{x}}{n+1}\right)$
$\Rightarrow \frac{y}{Q}=\frac{y}{x}+(r-1)\cdot \frac{(x-y)}{x(n+1)}$...(4)
Hence, $\left(\frac{P}{x}+\frac{y}{Q}\right)=1+\frac{y}{x}$ (independent of $n$ and $r$ )
[Put $x=5$ and $y=100]$
Here, $\left(\frac{P}{5}+\frac{100}{Q}\right)=1+\frac{100}{5}=21$. Ans.
Objective approach: Take $P$ as first arithmetic mean of 5 and 100 i.e. $P=\frac{105}{2}$.
Hence $\frac{P}{5}=\frac{21}{2}$ and $Q$ as first harmonic mean of 5 and 100 hence $Q=\frac{2\times 5\times 100}{105}$
Hence $\frac{100}{Q}=\frac{21}{2}$
$\therefore \left(\frac{P}{5}+\frac{100}{Q}\right)=21$