Question

If $f\left(n + 1\right)=\frac{2 f \left(n\right) + 1}{2}$ for $n=1,2,3\ldots \ldots ..$ and $f\left(1\right)=2$ , then $\frac{f \left(101\right)}{10}$ is equal to

Answer 1

$f\left(n + 1\right)=f\left(n\right)+\frac{1}{2}$
$\Rightarrow f\left(n + 1\right)-f\left(n\right)=\frac{1}{2}$
$\Rightarrow f\left(1\right),f\left(2\right),f\left(3\right)\ldots \ldots ..$ are in AP with common difference $\frac{1}{2}$ .
$f\left(101\right)=f\left(1\right)+100d$
$=2+100\times \frac{1}{2}=52$
$\Rightarrow \frac{f \left(101\right)}{10}=5.2$