Solution

Given, $$a_n= \dfrac {2n-3}{6}$$
Substituting $$n= 1, 2, 3, 4, 5$$, we obtain
$$\displaystyle { a }_{ 1

}=\frac { 2\times 1-3 }{ 6 } =\frac { -1 }{ 6 } \\\displaystyle { a }_{ 2 }=\frac {

2\times 2-3 }{ 6 } =\frac { 1 }{ 6 } \\ \displaystyle{ a }_{ 3 }=\frac { 2\times 3-3

}{ 6 } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \\ \displaystyle{ a }_{ 4 }=\frac {

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

}{ 6 } =\frac { 7 }{ 6 }$$