Question

Let $f\left(x\right)=(a^2+b2-4a-6b+13)(2x2-4x+5),a,b\in R$ such that $f\left(0\right)=f\left(1\right)=f\left(2\right)$. If $a,A_1,A_2..................A_{1}0$, $b$ is an $A.P$. and $a,H_1,H_2,.....................H_{1}0$, $b$ is a $H.P.$,
then find the value of $\frac{1}{10}\left(\stackrel{8}{\sum _{r=4}}{A}_r{H}_{11-r}\right)$

Answer 1

$f\left(x\right)=(a^2+b^2-4a-6b+13)(2x^2-4x+5)$

$f\left(0\right)=5(a^2+b^2-4a-6b+13)$

$f\left(1\right)=3(a^2+b^2-4a-6b+13)$

$f\left(2\right)=5(a^2+b^2-4a-6b+13)$

$f\left(0\right)=f\left(1\right)=f\left(2\right)$

$\Rightarrow a^2+b^2-4a-6b+13=0$.

$\Rightarrow a^2-4a+4+b^2-6b+9=0$.

$\Rightarrow (a-2{)}^2+(b-3{)}^2=0$

We know $a^2+b^2=0$ if and

only if $a=0,b=0$

$\therefore a-2=0$ and $b-3=0$

$\Rightarrow a=2,b=3$

Now

$a,A_1,....,A_{10},b$ are in $A.P$

Sp $b=a+11d$

$\Rightarrow 3=2+11d$

$\Rightarrow d=\frac{1}{11}$

$\therefore A_4=a+4d=2+\frac{4}{11}=\frac{26}{11}$

$A_5=a+5d=\frac{27}{11}$

$A_6=a+6d=\frac{28}{11}$

$A_7=a+7d=\frac{29}{11}$

$A_8=a+8d=\frac{30}{11}$

Again $a,H_1,....H_{10}$ we know

$\Rightarrow \frac{1}{a},\frac{1}{H_1},......,\frac{1}{H_{10}},\frac{1}{b}$ are in $A.P$

So $\frac{1}{b}=\frac{1}{a}+11d$

$\Rightarrow \frac{1}{3}=\frac{1}{2}+11d\Rightarrow d=\frac{-1}{66}$

So

$\frac{1}{H_3}=\frac{1}{a}+3d=\frac{1}{2}-\frac{3}{66}=\frac{30}{66}$

$H_3=\frac{66}{30}$

$\frac{1}{H_4}=\frac{1}{a}+4d=\frac{29}{66}\Rightarrow H_9=\frac{66}{29}$

$\frac{1}{H_5}=\frac{28}{66}\Rightarrow H_5=\frac{66}{28}$

$H_6=\frac{66}{27},H_7=\frac{66}{26}$

So $A_4{H}_7+A_5{H}_6+A_6{H}_5+A_7{H}_9+A_8{H}_3$

$=\frac{26}{11}\times \frac{66}{26}+\frac{27}{11}\times \frac{66}{27}+\frac{28}{11}\times \frac{66}{28}+\frac{29}{11}\times \frac{66}{29}+\frac{30}{11}\times \frac{66}{30}$

$=6+6+6+6+6=30$

$\therefore \frac{1}{10}\left({\sum }_{r=9}^8{A}_r{H}_{11-r}\right)=\frac{30}{10}=3$