An arithmetic progressions or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solutions:

(iii) ${{a}_{n}}={{3}^{n}}$

Given sequence whose ${{a}_{n}}={{3}^{n}}$

To get the first five terms of given sequence, put $n=1,2,3,4,5$ in the above

${{a}_{1}}={{3}^{1}}=3;$

${{a}_{2}}={{3}^{2}}=9;$

${{a}_{3}}=27;$

${{a}_{4}}={{3}^{4}}=81;$

${{a}_{5}}={{3}^{5}}=243.$

$\therefore $ the required first five terms of the sequence whose ${{n}^{th}}$ term, ${{a}_{n}}={{3}^{n}}$ are $3,9,27,81,243.$

${{a}_{n}}=\frac{3n-2}{5}$

.(iv) ${{a}_{n}}=\left( 3n-2 \right)/5$

Given sequence whose

To get the first five terms of the sequence, put $n=1,2,3,4,5$ in the above

And, we get

${{a}_{1}}=\frac{3\times 1-2}{5}=\frac{3-2}{5}=\frac{1}{5}$

${{a}_{2}}=\frac{3\times 2-2}{5}=\frac{6-2}{5}=\frac{4}{5}$

${{a}_{3}}=\frac{3\times 3-2}{5}=\frac{9-2}{5}=\frac{7}{5}$

${{a}_{4}}=\frac{3\times 4-2}{5}=\frac{12-2}{5}=\frac{10}{5}$

${{a}_{5}}=\frac{3\times 5-2}{5}=\frac{15-2}{5}=\frac{13}{5}$

$\therefore $ the required first five terms of the sequence are $1/5,4/5,7/5,10/5,13/5$