Let the arithmetic mean of (1)/(a) and (1)/(b) be (5)/(16), a2.

Question

Let the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ be $\frac{5}{16}$, $a>2$. If $\alpha $ is such that $a$, $4$, $\alpha $, $b$ are in A.P., then the equation $\alpha {x}^{2}-ax+2(\alpha -2b)=0$ has :

Accepted Answer

The arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ is $\frac{5}{16}$.

$\frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{5}{16}$

$\Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{5}{8}$

$\Rightarrow$ $\frac{a+b}{ab}=\frac{5}{8}$

$\Rightarrow 8(a+b)=5ab\, \, \, \, \, \, \, \, \, \, ...(i)$

$a,4,\alpha ,b$ are in A.P. Let the common difference be $d$.

(A). $4=a+d\implies d=4-a$

(B). $\alpha =a+2d=a+2(4-a)=8-a$

(C). $b=a+3d=a+3(4-a)=12-2a$

Solve for $a,b$, and $\alpha $

Substitute $b=12-2a$ into equation (1) :

$\begin{gathered} 8(a+12-2a)=5a(12-2a) \ 8(12-a)=60a-10{a}^{2} \ 96-8a=60a-10{a}^{2} \ 10{a}^{2}-68a+96=0 \ 5{a}^{2}-34a+48=0 \end{gathered}$

Using the quadratic formula: $a=\frac{34\pm \sqrt{{34}^{2}-4(5)(48)}}{2(5)}=\frac{34\pm \sqrt{1156-960}}{10}=\frac{34\pm 14}{10}$

$a=\frac{48}{10}=4.8,\, a=\frac{20}{10}=2a>2$

With $a=4.8:\, \alpha =8-4.8=3.2,\, b=12-2(4.8)=12-9.6=2.4$

Analyze the Quadratic Equation

The given equation is $\alpha {x}^{2}-ax+2(\alpha -2b)=0$. Substitute the values:

$\begin{gathered} 3.2{x}^{2}-4.8x+2(3.2-2(2.4))=0 \ 3.2{x}^{2}-4.8x+2(3.2-4.8)=0 \ 3.2{x}^{2}-4.8x-3.2=0 \ 2{x}^{2}-3x-2=0 \end{gathered}$

Factorise $(2x+1)(x-2)=0$

The roots are $x=2$ and $x=-0.5$.

Looking at the options: Root $x=2$ lies in the interval $(1,4)$. Root $x=-0.5$ lies in the interval $(-2,0)$.

Solutions · 1