MCQ
If p, q be two A.M.'s and G be one G.M. between two numbers, then G2 =
- A$(2\text{p}-\text{q})(\text{p}-2\text{q})$
- B$(2\text{p}-\text{q})(2\text{q}-\text{p})$
- C$(2\text{p}-\text{q})(\text{p}+2\text{q})$
- DNone of these.
Solution:
Let the two numbers be a and b.
a, p, q and b are in A.P.
$\therefore\text{ p}-\text{a}=\text{q}-\text{q}=\text{b}-\text{q}$
$\Rightarrow\text{ p}-\text{a}=\text{q}-\text{p}\text{ and}\text{ q}-\text{p}=\text{b}-\text{q}$
$\Rightarrow\text{ a}=2\text{p}-\text{q}\text{ and}\text{ b}=2\text{q}-\text{p}\cdots(\text{i})$
Also, a, G and b are in G.P.
$\therefore\text{G}^2=\text{ab}$
$\Rightarrow\text{G}^2=(2\text{p}-\text{q})(2\text{q}-\text{p})$
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The equation of the locus of a point equidistant from the point A (1, 3) and B (-2, 1) is:
If slope of a line is 4 and x-intercept made by the line is 2 then the equation of line will be:
The equations of the lines passing through the point (1, 0) and at a distance $\frac{\sqrt{3}}{2}$ from the origin, are
$\sqrt{3}\text{x}+\text{y}-\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0$
$\sqrt{3}\text{x}+\text{y}+\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}+\sqrt{3}=0$
$\text{x}+\sqrt{3}\text{y}-\sqrt{3}=0,\text{x}-\sqrt{3}\text{y}-\sqrt{3}=0$
None of these.