Write the first three terms of the APs when $ a $ and $ d $ are as given below:
$ a=\frac{1}{2}, d=-\frac{1}{6} $

To do:

We have to write the first three terms of the given arithmetic progressions. 

Solution:

(i) First term $a_1=a=\frac{1}{2}$

Second term $a_2=a_1+d=\frac{1}{2}+\frac{-1}{6}=\frac{1\times3-1}{6}=\frac{2}{6}=\frac{1}{3}$

Third term $a_3=a_2+d=\frac{1}{3}+\frac{-1}{6}=\frac{1\times2-1}{6}=\frac{1}{6}$

Therefore, the first three terms of the given AP are $\frac{1}{2}, \frac{1}{3}, \frac{1}{6}$. 

(ii) First term $a_1=a=-5$

Second term $a_2=a_1+d=-5+(-3)=-5-3=-8$

Third term $a_3=a_2+d=-8+(-3)=-8-3=-11$

Therefore, the first three terms of the given AP are $-5, -8, -11$. 

(iii) First term $a_1=a=\sqrt{2}$

Second term $a_2=a_1+d=\sqrt{2}+\frac{1}{\sqrt{2}}=\frac{\sqrt2 \times \sqrt2+1}{\sqrt2}=\frac{2+1}{\sqrt{2}}=\frac{3}{\sqrt{2}}$

Third term $a_3=a_2+d=\sqrt{2}+\frac{2}{\sqrt{2}}=\frac{\sqrt2 \times \sqrt2+2}{\sqrt2}=\frac{2+2}{\sqrt{2}}=\frac{4}{\sqrt{2}}$

Therefore, the first three terms of the given AP are $\sqrt{2}, \frac{3}{\sqrt{2}}, \frac{4}{\sqrt{2}}$.