The roots of equation $x^{2} + 2 (a + 3) x + 9 = 0$ lie between $- 6$ and $1$ and $2$ , $h_{1}, h_{2}, . . . h_{20} [a]$ are in H.P., where $[a]$ denotes the integral part of $a$ , and $2$ , $a_{1}, a_{2}, . . ., a_{20}, [a]$ are in A.P. then $a_{3} h_{18} =$

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Solution

Since roots of equation $f (x) = x^{2} + 2 (a - 3) + 9 = 0$ lie between $- 6$ and $1$
$∴$ (i) $D \geq 0$
(ii) $f (- 6) > 0$
(iii) $f (1) > 0$
(iv) $- 6 < \frac{α + β}{2}$
(v) $1 > \frac{α + β}{2}$
Hence $6 \leq a \leq \frac{27}{4}$
$∴ [a] = 6, a_{3} = 2 + 3 d = 2 + 3. \frac{6 - 2}{21} = 2 + \frac{4}{7} = \frac{18}{7}$
$\frac{1}{h_{18}} = \frac{1}{2} + 18. ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{1}{6} - \frac{1}{2}}{21} ⎞ ⎟ ⎟ ⎟ ⎠ = \frac{1}{2} - \frac{2}{7} = \frac{3}{14}$
$∴ a_{3} h_{18} = \frac{18}{7} . \frac{14}{3} = 12$

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The roots of the equation $x^{2} + 2 (a - 3) x + 9 = 0$ lie between $- 6$ and $1$ and $2, h_{1}, h_{2}, . . . . h_{20}, [a]$ are in H.P. and $2, a_{1}, a_{2} . . . . a_{20}, [a]$ are in A.P. where $[a]$ denotes the integral part of $a$ , then let the value of $a_{3} h_{18}$ be equal to $k$ then, find the sum of the digits of $k$ is equal to