Question

If [x] denotes the integral part of x for real x, then the value of
$\left[\frac{1}{4}\right]+[\frac{1}{4}+\frac{1}{200}]+[\frac{1}{4}+\frac{1}{100}]+[\frac{1}{4}+\frac{3}{200}]+.......+[\frac{1}{4}+\frac{199}{200}]$ is 10a. Find a.

Answer 1

$\left[\frac{1}{4}\right]+[\frac{1}{4}+\frac{1}{200}]+[\frac{1}{4}+\frac{1}{100}]+[\frac{1}{4}+\frac{3}{200}]+.......+[\frac{1}{4}+\frac{199}{200}]$
We know that
$0\le x<1$ , $\left[x\right]=0$
and $1\le x<2$, $\left[x\right]=1$
Now, $[\frac{1}{4}+\frac{3}{4}]=[\frac{1}{4}+\frac{150}{200}]=\left[1\right]=1$
Similarly, $[\frac{1}{4}+\frac{151}{200}]=\left[1.005\right]=1$
So, we can write the nth term of the series in the form $[\frac{1}{4}+\frac{n}{200}]$
Here, when $n<150$, $[\frac{1}{4}+\frac{n}{200}]=0$
When $150\le n\le 199$ , $[\frac{1}{4}+\frac{n}{200}]=1$.

So, $\left[\frac{1}{4}\right]+[\frac{1}{4}+\frac{1}{200}]+[\frac{1}{4}+\frac{1}{100}]+[\frac{1}{4}+\frac{3}{200}]+.......+[\frac{1}{4}+\frac{150}{200}]......+[\frac{1}{4}+\frac{199}{200}]$
$=0+0+....+1+1+1+....upto50times$
$=50$
But given sum is $10a$
$\Rightarrow 10a=50$
$\Rightarrow a=5$