The equation whose roots are the values of $r$ satisfying the equation $^{69} C_{3 r - 1} -^{69} C_{r^{2}} =^{69} C_{r^{2} - 1} -^{69} C_{3 r}$ is

x^{3} - 10 x^{2} + 21 x = 0

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x^{2} + 10 x + 21 = 0

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x^{2} - 10 x + 21 = 0

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x^{2} - 21 x + 10 = 0

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Solution

The correct option is C $x^{2} - 10 x + 21 = 0$
Given, $^{69} C_{3 r - 1} -^{69} C_{r^{2}} =^{69} C_{r^{2} - 1} -^{69} C_{3 r}$
$\Rightarrow^{69} C_{3 r} +^{69} C_{3 r - 1} =^{69} C_{r^{2}} +^{69} C_{r^{2} - 1}$
$\Rightarrow^{70} C_{3 r} =^{70} C_{r^{2}} (∵^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r})$
$\Rightarrow 3 r = r^{2} or 3 r + r^{2} = 70$
$\Rightarrow r (r - 3) = 0 or (r - 7) (r + 10) = 0$
$\Rightarrow r = 0, 3 or r = 7, - 10$

But for $r = 0, - 10;^{69} C_{3 r - 1}$ is not defined.
$∴ r = 3, 7$
Equation whose roots are $3, 7$ is
$x^{2} - (3 + 7) x + 3 \cdot 7 = 0$
$\Rightarrow x^{2} - 10 x + 21 = 0$

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The equation whose roots are the values of r satisfying the equation 69C3r−1−69Cr2=69Cr2−1−69C3r is

The equation whose roots are the values of $r$ satisfying the equation $^{69} C_{3 r - 1} -^{69} C_{r^{2}} =^{69} C_{r^{2} - 1} -^{69} C_{3 r}$ is