STD 11 - 5. linear inequalities — Maths STD 11 Science — Question

We have,

$\Rightarrow \sqrt{n+1}-\sqrt{n-1} < 0.2 n \in N$

$\sqrt{n+1} < 0.2+\sqrt{n-1}$

On squaring both side, we get

$n+1 < 0.04+n-1+0.4 \sqrt{n-1}$

$\Rightarrow n+1-n+1-0.04 < 0.4 \sqrt{n-1}$

$\Rightarrow \frac{2-0.04}{0.4} < \sqrt{n-1}$

$\Rightarrow 49 < \sqrt{n-1}$

$\Rightarrow n-1 > (49)^2$

$\Rightarrow n > 1+2401$

$\Rightarrow n > 25.01$

$\therefore \text { Minimum value of } n=26$