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

Let $x =5$ and $y =100$.
Now, $x, A_1, A_2, A_3, \ldots \ldots . . ., A_n$, $y$ are in A.P. ....(1)
and $ x, H_1, H_2, H_2, H_3, \ldots, 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),
$\frac{1}{ x }, \frac{1}{ H _1}, \frac{1}{ H _2}, \ldots \ldots \ldots, \frac{1}{ H _{ n }}, \frac{1}{ y }$ are in A.P.
$\therefore \frac{1}{y}=(n+2)^{\text {th }} \operatorname{term}=\frac{1}{x}+(n+1) d^{\prime} \Rightarrow d^{\prime}=\left(\frac{\frac{1}{y}-\frac{1}{x}}{n+1}\right) $
$\therefore \frac{1}{Q}=\frac{1}{H_{r-1}}=r^{\text {th }} \operatorname{term}=\frac{1}{x}+(r-1) d^{\prime}=\frac{1}{x}+(r-1)\left(\frac{\frac{1}{y}-\frac{1}{x}}{n+1}\right)$
$\Rightarrow \frac{ y }{ Q }=\frac{ y }{ x }+( r -1) \cdot \frac{( x - y )}{ x ( n +1)}$...(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 $\frac{100}{Q}=\frac{21}{2}$
$\therefore \left(\frac{ P }{5}+\frac{100}{ Q }\right)=21$