Question

The roots of equation $x^2+2(a+3)x+9=0$ lie between $-6$ and $1$ and $2$,$h_1,h_2,...h_{20}\left[a\right]$are in H.P., where $\left[a\right]$ denotes the integral part of $a$, and $2$,$a_1,a_2,...,a_{20},\left[a\right]$ are in A.P. then $a_3{h}_{18}=$

Answer 1

Since roots of equation $f\left(x\right)=x^2+2(a-3)+9=0$ lie between $-6$ and $1$
$\therefore$ (i) $D\ge 0$
(ii) $f(-6)>0$
(iii) $f\left(1\right)>0$
(iv) $-6<\frac{\alpha +\beta }{2}$
(v) $1>\frac{\alpha +\beta }{2}$
Hence $6\le a\le \frac{27}{4}$
$\therefore \left[a\right]=6,a_3=2+3d=2+3.\frac{6-2}{21}=2+\frac{4}{7}=\frac{18}{7}$
$\frac{1}{h_{18}}=\frac{1}{2}+18.\left(\frac{\frac{1}{6}-\frac{1}{2}}{21}\right)=\frac{1}{2}-\frac{2}{7}=\frac{3}{14}$
$\therefore a_3{h}_{18}=\frac{18}{7}.\frac{14}{3}=12$