To find $ S_{ 20 } $ we use formula

$$ \color{blue}{S_n = \frac{n}{2} \cdot \left(2a_1 + (n-1) \cdot d \right) } $$

In this example we have $ a_1 = 5 ~,~ d = 3 ~,~ n = 20 $. After substituting these values into the above equation, we obtain:

$$ \begin{aligned} S_n &= \frac{n}{2} \cdot \left(2a_1 + (n-1) \cdot d \right) \\[1 em] S_{ 20 } &= \frac{ 20 }{2} \cdot \left( 2 \cdot 5 + ( 20-1) \cdot 3 \right) \\[1 em] S_{ 20 } &= \frac{ 20 }{2} \cdot \left( 10 + 19 \cdot 3 \right) \\[1 em] S_{ 20 } &= \frac{ 20 }{2} \cdot \left( 10 + 57 \right) \\[1 em] S_{ 20 } &= \frac{ 20 }{2} \cdot 67 \\[1 em] S_{ 20 } &= 670 \end{aligned}$$

The first few terms of this sequence are:

$$ 5, ~~~8, ~~~11, ~~~14, ~~~17 . . . $$

Arithmetic sequences