To find $ S_{ 10 } $ 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 = 4 ~,~ d = 5 ~,~ n = 10 $. After substituting these values into the above equation, we obtain:

$$ \begin{aligned} S_n &= \frac{n}{2} \cdot \left(2a_1 + (n-1) \cdot d \right) \\[1 em] S_{ 10 } &= \frac{ 10 }{2} \cdot \left( 2 \cdot 4 + ( 10-1) \cdot 5 \right) \\[1 em] S_{ 10 } &= \frac{ 10 }{2} \cdot \left( 8 + 9 \cdot 5 \right) \\[1 em] S_{ 10 } &= \frac{ 10 }{2} \cdot \left( 8 + 45 \right) \\[1 em] S_{ 10 } &= \frac{ 10 }{2} \cdot 53 \\[1 em] S_{ 10 } &= 265 \end{aligned}$$

The first few terms of this sequence are:

$$ 4, ~~~9, ~~~14, ~~~19, ~~~24 . . . $$

Arithmetic sequences