1)

The value \(\displaystyle{u}_{{1}}={2}\)

Common difference= 3

use the formula

\(\displaystyle{u}_{{n}}={u}_{{1}}+{\left({n}-{1}\right)}{d}\)

Replace \(\displaystyle{u}_{{1}}={2},\ {d}={3}\)

\(\displaystyle\Rightarrow{u}_{{n}}={2}+{\left({n}+{1}\right)}{3}\)

\(\displaystyle\Rightarrow{3}{n}-{3}+{2}\)

\(\displaystyle\Rightarrow{u}_{{n}}={3}{n}-{1}\) \(\displaystyle{n}={1}\ {u}_{{1}}={2}\)

\(\displaystyle{n}={2}\ {u}_{{2}}={5}\)

\(\displaystyle{n}={3}\ {u}_{{3}}={8}\)

\(\displaystyle{n}={4}\ {u}_{{4}}={11}\)

\(\displaystyle{n}={5}\ {u}_{{5}}={14}\)

\(\displaystyle{n}={6}\ {u}_{{6}}={17}\)

2)

a= first term= 27

\(\displaystyle\gamma\)= common ration= \(\displaystyle{\frac{{{1}}}{{{3}}}}\)

We know the formula

\(\displaystyle{a}_{{n}}={a}\gamma^{{{n}-{1}}}\)

Replace a=27, \(\displaystyle\gamma={\frac{{{1}}}{{{3}}}}\)

\(\displaystyle{a}_{{n}}={27}{\left({\frac{{{1}}}{{{3}}}}\right)}^{{{n}-{1}}}\)

\(\displaystyle{a}_{{n}}={\left({3}\right)}^{{{3}}}{\left({\frac{{{1}}}{{{3}}}}\right)}^{{{n}-{1}}}\) \(\displaystyle{a}_{{n}}={3}^{{{3}}}\cdot{3}^{{-{\left({n}-{1}\right)}}}\)

\(\displaystyle{a}_{{n}}={3}^{{{4}-{n}}}\) \(\displaystyle{a}_{{1}}={27},\ {a}_{{2}}={9},\ {a}_{{3}}={3},\ {a}_{{4}}={0},\ {a}_{{5}}={\frac{{{1}}}{{{3}}}},{a}_{{6}}={\frac{{{1}}}{{{9}}}}\)

The value \(\displaystyle{u}_{{1}}={2}\)

Common difference= 3

use the formula

\(\displaystyle{u}_{{n}}={u}_{{1}}+{\left({n}-{1}\right)}{d}\)

Replace \(\displaystyle{u}_{{1}}={2},\ {d}={3}\)

\(\displaystyle\Rightarrow{u}_{{n}}={2}+{\left({n}+{1}\right)}{3}\)

\(\displaystyle\Rightarrow{3}{n}-{3}+{2}\)

\(\displaystyle\Rightarrow{u}_{{n}}={3}{n}-{1}\) \(\displaystyle{n}={1}\ {u}_{{1}}={2}\)

\(\displaystyle{n}={2}\ {u}_{{2}}={5}\)

\(\displaystyle{n}={3}\ {u}_{{3}}={8}\)

\(\displaystyle{n}={4}\ {u}_{{4}}={11}\)

\(\displaystyle{n}={5}\ {u}_{{5}}={14}\)

\(\displaystyle{n}={6}\ {u}_{{6}}={17}\)

2)

a= first term= 27

\(\displaystyle\gamma\)= common ration= \(\displaystyle{\frac{{{1}}}{{{3}}}}\)

We know the formula

\(\displaystyle{a}_{{n}}={a}\gamma^{{{n}-{1}}}\)

Replace a=27, \(\displaystyle\gamma={\frac{{{1}}}{{{3}}}}\)

\(\displaystyle{a}_{{n}}={27}{\left({\frac{{{1}}}{{{3}}}}\right)}^{{{n}-{1}}}\)

\(\displaystyle{a}_{{n}}={\left({3}\right)}^{{{3}}}{\left({\frac{{{1}}}{{{3}}}}\right)}^{{{n}-{1}}}\) \(\displaystyle{a}_{{n}}={3}^{{{3}}}\cdot{3}^{{-{\left({n}-{1}\right)}}}\)

\(\displaystyle{a}_{{n}}={3}^{{{4}-{n}}}\) \(\displaystyle{a}_{{1}}={27},\ {a}_{{2}}={9},\ {a}_{{3}}={3},\ {a}_{{4}}={0},\ {a}_{{5}}={\frac{{{1}}}{{{3}}}},{a}_{{6}}={\frac{{{1}}}{{{9}}}}\)