Question

If $f\left(x\right)$ be a function such that $f(x-1)+f(x+1)=\sqrt{3}f\left(x\right)$ and $f\left(5\right)=10,$ then the sum of digits of the value of $\sum _{r=0}^{19}f(5+12r)$ is ......

Answer 1

$f(x-1)+f(x+1)=\sqrt{3}f\left(x\right)$
$\Rightarrow f\left(x\right)+f(x+2)=\sqrt{3}f(x+1)$
Putting, $x=x+2$,
$f(x+1)+f(x+3)=\sqrt{3}f(x+2)$
$f(x-1)+2f(x+2)+f(x+3)=\sqrt{3}\left[\sqrt{3}f\right(x+1\left)\right]$
$f(x-1)+f(x+3)=f(x+1)$
Putting, $x=x+2$, again
$f(x+1)+f(x+5)=f(x+3)$
$f(x-1)+f(x+5)=0$
$f(x+5)=-f(x-1)$
$f\left(x\right)=-f(x+6)$
$f(x+12)=f\left(x\right)$
${\sum }_{r-0}^{19}f(5+12r)=20f\left(5\right)=20\times 10=200$
Hence, sum of digits $=2+0+0=2$