Question

The roots of the equation $x^2+2(a-3)x+9=0$ lie between $-6$ and $1$ and $2,h_1,h_2,....h_{20},\left[a\right]$ are in H.P. and $2,a_1,a_2....a_{20},\left[a\right]$ are in A.P. where $\left[a\right]$ 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

Answer 1

The root of the equation ................................
$6\le a<6.25\Rightarrow \left[a\right]=6$
$\Rightarrow 2,h_1.....h_{20},6\to H.P.$
$h_{18}=\frac{14}{3}$
$2,a_1.....a_{20},6\to A.Pv\Rightarrow a_3=\frac{18}{7}$
$a_3{h}_{18}=12$

This shows that the value of $k$ is $12$

Sum of the digits of $k$ is $1+2=3$