Question

$x+y+z=15$ when $a,x,y,z,b$ are in AP and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3}$ when a, x, y, z, b are in HP then the quadratic equation whose roots are $\frac{1}{a}and\frac{1}{b}$ is

• A. $x^2-10x+9=0$
• B. $x^2+10x-9=0$
• C. $9x^2-10x+1=0$
• D. $9x^2+10x+1=0$

Answer 1

The correct option is C $9x^2-10x+1=0$

if $a,x,y,z,b$ are in AP.

$a+x+y+z+b=\frac{5}{2}(a+b)$

$(a+b+15)\times 2=5(a+b)$

$3(a+b)=30$

$(a+b)=10$

if $,x,y,z,b$ are in HP then their reciprocal will be in AP

$\frac{1}{a}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{b}=\frac{5}{2}(\frac{1}{a}+\frac{1}{b})$

$(\frac{1}{a}+\frac{1}{b}+\frac{5}{3})\times 2=5\times \left(\frac{1}{a}+\frac{1}{b}\right)$

$\frac{10}{3}=\frac{3(a+b)}{ab}$
$ab=9$ (from eq (1) putting value of $(a+b)$)

if roots are $\frac{1}{a},\frac{1}{b}$
sum of roots $=\frac{a+b}{ab}$

$=\frac{10}{9}$

product of roots $=\frac{1}{ab}$

$=\frac{1}{9}$

quadratic eq:

$x^2-\left(\text{sum of the roots}\right)x+\text{product of the roots}=0$
$x^2-\frac{10}{9}x+\frac{1}{9}=0$

$9x^2-10x+1=0$