JEE MAIN - Mathematics (2003 - No. 36)
If the sum of the roots of the quadratic equation $$a{x^2} + bx + c = 0$$ is equal to the sum of the squares of their reciprocals, then $${a \over c},\,{b \over a}$$ and $${c \over b}$$ are in
Arithmetic - Geometric Progression
Arithmetic Progression
Geometric Progression
Harmonic Progression
Explanation
$$a{x^2} + bx + c = 0,$$ $$\alpha + \beta = {{ - b} \over a},\alpha \beta = {c \over a}$$
As for given condition, $$\alpha + \beta = {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}}$$
$$\alpha + \beta = {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} - {b \over a}$$
$$ = {{{{{b^2}} \over {{a^2}}} - {{2c} \over a}} \over {{{{c^2}} \over {{a^2}}}}}$$
On simplification $$2{a^2}c = a{b^2} + b{c^2}$$
$$ \Rightarrow {{2a} \over b} = {c \over a} + {b \over c}$$
$$ \Rightarrow {c \over a},{a \over b},{b \over c}$$ are in $$A.P.$$
$$\therefore$$ $${a \over c},{b \over a},\,\,\& \,\,$$ are in $$H.P.$$
As for given condition, $$\alpha + \beta = {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}}$$
$$\alpha + \beta = {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} - {b \over a}$$
$$ = {{{{{b^2}} \over {{a^2}}} - {{2c} \over a}} \over {{{{c^2}} \over {{a^2}}}}}$$
On simplification $$2{a^2}c = a{b^2} + b{c^2}$$
$$ \Rightarrow {{2a} \over b} = {c \over a} + {b \over c}$$
$$ \Rightarrow {c \over a},{a \over b},{b \over c}$$ are in $$A.P.$$
$$\therefore$$ $${a \over c},{b \over a},\,\,\& \,\,$$ are in $$H.P.$$
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