Question

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

Answer 1

Answer: 5.2

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