**Question:**

Write the first three terms of the AP's, when a and $d$ are as given below

(i) $a=\frac{1}{2}, d=\frac{-1}{6}$

(ii) $a=-5, d=-3$

(iii) $a=\sqrt{2}, d=\frac{1}{\sqrt{2}}$

**Solution:**

(i) Given that, first term $(a)=\frac{1}{2}$ and common difference $(d)=-\frac{1}{6}$

$\because$ nth term of an AP, $T_{n}=a+(n-1) d$

$\therefore$ Second term of an AP, $T_{2}=a+d=\frac{1}{2}-\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$

and third term of an AP, $T_{3}=a+2 d=\frac{1}{2}-\frac{2}{6}=\frac{1}{2}-\frac{1}{3}=\frac{3-2}{6}=\frac{1}{6}$

Hence, required three terms are $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}$.

(ii) Given that, first term $(a)=-5$ and common difference $(d)=-3$

$\because n$th term of an AP, $T_{n}=a+(n-1) d$

$\therefore$ Second term of an AP, $T_{2}=a+d=-5-3=-8$

and third term of an $\mathrm{AP}, T_{3}=\mathrm{a}+2 d=-5+2(-3)$

$=-5-6=-11$

Hence, required three terms are $-5,-8,-11$

(iii) Given that, first term $(a)=\sqrt{2}$ and common difference $(d)=\frac{1}{\sqrt{2}}$

$\because n$th term of an AP, $T_{n}=a+(n-1) d$

$\therefore$ Second term of an AP, $T_{2}=a+d=\sqrt{2}+\frac{1}{\sqrt{2}}=\frac{2+1}{\sqrt{2}}=\frac{3}{\sqrt{2}}$

and third term of an AP, $T_{3}+a+2 d=\sqrt{2}+\frac{2}{\sqrt{2}}=\frac{2+2}{\sqrt{2}}=\frac{4}{\sqrt{2}}$

Hence, required three terms are $\sqrt{2}, \frac{3}{\sqrt{2}}, \frac{4}{\sqrt{2}}$.