To find $ a_{ 68 } $ we use formula

$$ \color{blue}{a_n = a_1 + (n-1)d}$$

In this example we have $ a_1 = 1,~~ d = 3 ~~,~~ n = 68 $. After substituting these values into the formula, we obtain:

$$ \begin{aligned} a_n &= a_1 + d(n-1) \\[1 em] a_{ 68 } &= 1 + (68-1) \cdot 3 \\[1 em] a_{ 68 } &= 1 + 201 \\[1 em] a_{ 68 } &= 202 \end{aligned} $$

The first few terms of this sequence are:

$$ 1, ~~~4, ~~~7, ~~~10, ~~~13 . . . $$

Arithmetic sequences