Question

If $a,a_1,a_2,a_3,\dots \dots ,a_{2n-1},b$ are in AP, a, $b_1,b_2,b_3,\dots \dots b_{2n-1},b$ are in GP and $a,c_1,c_2,c_3,\dots .c_{2n-1},b$ are in $HP$, 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)

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