Sequences And Series Ques 25
- If $p$ is the first of the $n$ arithmetic means between two numbers and $q$ be the first on $n$ harmonic means between the same numbers. Then, show that $q$ does not lie between $p$ and $\left(\frac{n+1}{n-1}\right)^2 p$.
(1991, 4M)
Show Answer
Solution: Let two numbers be $a$ and $b$ and $A_1, A_2, \ldots, A_n$ be $n$ arithmetic means between $a$ and $b$.
Then, $a, A_1, A_2, \ldots, A_n, b$ are in AP with common difference
$ d =\frac{b-a}{n+1} $
$ p =A_1=a+d=a+\frac{b-a}{n+1}$
$\Rightarrow \quad p =\frac{n a+b}{n+1}$ $\quad$ ……..(i)
Let $H_1, H_2, \ldots, H_n$ be $n$ harmonic means between $a$ and $b$.
$\therefore \quad \frac{1}{a}, \frac{1}{H_1}, \frac{1}{H_2}, \ldots, \frac{1}{H_n}, \frac{1}{b}$ is an AP with common difference, $D=\frac{(a-b)}{(n+1) a b}$.
$\therefore \quad \frac{1}{q}=\frac{1}{a}+D \Rightarrow \frac{1}{q}=\frac{1}{a}+\frac{(a-b)}{(n+1) a b}$
$\Rightarrow \quad \frac{1}{q}=\frac{n b+a}{(n+1) a b}$
$\Rightarrow \quad q=\frac{(n+1) a b}{n b+a}$ $\quad$ ……..(ii)
From Eq. (i),
$ b=(n+1) p-n a . $
Putting it in Eq. (ii), we get
$\quad q\left\{n(n+1) p-n^2 a+a\right\}=(n+1) a\{(n+1) p-n a\} $
$\Rightarrow \quad n(n+1) a^2-\left\{(n+1)^2 p+\left(n^2-1\right) q\right\} a + n (n+1 ) pq = 0$
$\Rightarrow \quad n a^2-\{(n+1) p+(n-1) q\} a+n p q=0$
Since, $a$ is real, therefore
$ \begin{array}{lrl} \quad\quad \{(n+1) p+(n-1) q\}^2-4 n^2 p q>0 \\ \Rightarrow \quad (n+1)^2 p^2+(n-1)^2 q^2+2\left(n^2-1\right) p q-4 n^2 p q>0 \\ \Rightarrow \quad (n+1)^2 p^2+(n-1)^2 q^2-2\left(n^2+1\right) p q>0 \\ \Rightarrow \quad q^2-\frac{2\left(n^2+1\right)}{(n-1)^2} p q+\left(\frac{n+1}{n-1}\right)^2 p^2>0 \\ \Rightarrow \quad q^2-\left\{1+\left(\frac{n+1}{n-1}\right)^2\right\} p q+\left(\frac{n+1)}{n-1}\right)^2 p^2>0 \\ \Rightarrow \quad(q-p)\left\{q-\left(\frac{n+1}{n-1}\right)^2 p\right\}>0 \end{array} $
$\Rightarrow \quad q<p$ or $q>\left(\frac{n+1}{n-1}\right)^2 p$
$\because \quad\left\{\left(\frac{n+1}{n-1}\right)^2 p>p\right\}$
Hence, $q$ cannot lie between $p$ and $\left(\frac{n+1}{n-1}\right)^2 p$.
BETA
