Question

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

The correct option is D none of these

$a$, $a_1$, $a_2$, $a_3$, ..., $a_{2n-1}$, $b$ are in AP, $a$, $b_1$, $b_2$, $b_3$, ..., $b_{2n-1}$, $b$ are in GP and $a$, $c_1$, $c_2$, $c_3$, ..., $c_{2n-1}$, $b$ are in HP
$\Rightarrow$ $a_n,b_n,c_n$ are $A.M,G.M,H.M$ of respective series.
$\therefore a_n=\frac{a+b}{2}$ and $b_n^2=ab$ and $c_n=\frac{2ab}{a+b}$
The given equation is $a_n{x}^2-b_{n}x+c_n=0$
$\Rightarrow D=b_n^2-4a_n{c}_n=-3ab<0$ ($\because a,b$ are positive)
Hence, roots are non-real (not imaginary) . Hence, option D is correct.