Question

If $a, a_1, a_2, a_3, \ldots \ldots, a_{2 n-1}, b$ are in AP, a, $b_1, b_2, b_3, \ldots \ldots b_{2 n-1}, b$ are in GP and $a, c_1, c_2, c_3, \ldots . c_{2 n-1}, b$ are in $H P$, where $a, b$ are positive, then the equation $a_n x^2-b_n x+c_n=0$ has its roots

• A. real and unequal
• B. real and equal
• C. imaginary
• D. None of these

Answer 1

Answer: (C)

$a, a_1, a_2 , a_3 \ldots . . a_{2 n-1} b$ are in A.P.
$a, b_1, b_2, b_3 \ldots \ldots . b_{2 n-1}, b$ are in G.P.
$a, c_1, c_2, c_3 \ldots . . c_{2 n-1}, b$ are in H.P.
There are $2 n+1$ terms in A.P., G.P. and H.P. If common difference is $d$ for A.P. then $d=\frac{b-a}{2 n}$
$\therefore a_n$ is $(n+1)^{\text {h }}$ term;
$\therefore a_n=a+\frac{n(b-a)}{2 n}=\frac{a+b}{2}$
If $r$ is common ratio for G.P. $b=a(r)^{2 n}$
$\therefore r=\left(\frac{b}{a}\right)^{\frac{1}{2 n}}$
$\therefore b_n=a^n=a\left(\frac{b}{a}\right)^{\frac{1}{2 n} \cdot n}$
$\therefore b_n=\sqrt{a b}$
similarly $c_n=\frac{2 a b}{a+b}$
in equation $a_n x^2-b_n x+c_n=0$
$D=b_n^2-4 a_n c_n=a b-4\left(\frac{a+b}{2}\right) \cdot \frac{2 a b}{a+b}=a b-4 a b=-3 a b$