To find $ S_{ 25 } $ 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 = 2 ~,~ d = 6 ~,~ n = 25 $. After substituting these values into the above equation, we obtain:

$$ \begin{aligned} S_n &= \frac{n}{2} \cdot \left(2a_1 + (n-1) \cdot d \right) \\[1 em] S_{ 25 } &= \frac{ 25 }{2} \cdot \left( 2 \cdot 2 + ( 25-1) \cdot 6 \right) \\[1 em] S_{ 25 } &= \frac{ 25 }{2} \cdot \left( 4 + 24 \cdot 6 \right) \\[1 em] S_{ 25 } &= \frac{ 25 }{2} \cdot \left( 4 + 144 \right) \\[1 em] S_{ 25 } &= \frac{ 25 }{2} \cdot 148 \\[1 em] S_{ 25 } &= 1850 \end{aligned}$$

The first few terms of this sequence are:

$$ 2, ~~~8, ~~~14, ~~~20, ~~~26 . . . $$

Arithmetic sequences