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 $\sqrt{2}, \frac{3}{\sqrt{2}}, \frac{4}{\sqrt{2}}$.
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