A problem by crazy singh

Level pending

If $x_{1}$ , $x_{2}$ , and $x_{3}$ are the abscissa of the points $A_{1}$ , $A_{2}$ and $A_{3}$ respectively where the lines $y=m_{1} x$ , $y=m_{2} x$ and $y=m_{3} x$ meet the line $2x-y+3=0$ such that $m_{1}$ , $m_{2}$ , and $m_{3}$ are in A.P., then $x_{1}$ , $x_{2}$ , and $x_{3}$ are in

A.P.
G.P.
H.P.
None of these

2 3 1 4

1 solution

Tom Engelsman
Dec 25, 2019

Let $m_{1} < m_{2} < m_{3}$ with $m_{2} = m_{1} + d, m_{3} = m_{1} + 2d$ for positive difference $d.$ Setting each of the above lines equal to the line $y = 2x+3$ yields:

$m_{1}x = 2x+3 \Rightarrow \boxed{x_{1} = \frac{3}{m_{1} -2}};$

$m_{2}x = (m_{1} + d)x = 2x+3 \Rightarrow \boxed{x_{2} = \frac{3}{(m_{1} - 2) + d}};$

$m_{3}x = (m_{1} + 2d)x = 2x+3 \Rightarrow \boxed{x_{3} = \frac{3}{(m_{1} - 2) + 2d}}.$

Thus $x_{1}, x_{2}, x_{3}$ are in harmonic progression (H.P.).

A problem by crazy singh

1 solution

0 pending reports