1)

The value \(u_1=2\)

Common difference= 3

use the formula

\(u_n=u_1+(n-1)d\)

Replace \(u_1=2, \ d=3\)

\(\Rightarrow u_n=2+(n+1)3\)

\(\Rightarrow 3n-3+2\)

\(\Rightarrow u_n=3n-1\) \(n=1 \ u_1=2\)

\(n=2 \ u_2=5\)

\(n=3 \ u_3=8\)

\(n=4 \ u_4=11\)

\(n=5 \ u_5=14\)

\(n=6 \ u_6=17\)

2)

a= first term= 27

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

We know the formula

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

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

\(a_n=27(\frac{1}{3})^{n-1}\)

\(a_n=(3)^{3}(\frac{1}{3})^{n-1}\) \(a_n=3^{3}\cdot 3^{-(n-1)}\)

\(a_n=3^{4-n}\) \(a_1=27, \ a_2=9, \ a_3=3, \ a_4=0, \ a_5=\frac{1}{3}, a_6=\frac{1}{9}\)

The value \(u_1=2\)

Common difference= 3

use the formula

\(u_n=u_1+(n-1)d\)

Replace \(u_1=2, \ d=3\)

\(\Rightarrow u_n=2+(n+1)3\)

\(\Rightarrow 3n-3+2\)

\(\Rightarrow u_n=3n-1\) \(n=1 \ u_1=2\)

\(n=2 \ u_2=5\)

\(n=3 \ u_3=8\)

\(n=4 \ u_4=11\)

\(n=5 \ u_5=14\)

\(n=6 \ u_6=17\)

2)

a= first term= 27

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

We know the formula

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

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

\(a_n=27(\frac{1}{3})^{n-1}\)

\(a_n=(3)^{3}(\frac{1}{3})^{n-1}\) \(a_n=3^{3}\cdot 3^{-(n-1)}\)

\(a_n=3^{4-n}\) \(a_1=27, \ a_2=9, \ a_3=3, \ a_4=0, \ a_5=\frac{1}{3}, a_6=\frac{1}{9}\)