To find $ S_{ 0 } $ 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 = \frac{ 1220 }{ 51 } ~,~ d = 0 ~,~ n = 0 $. After substituting these values into the above equation, we obtain:

$$ \begin{aligned} S_n &= \frac{n}{2} \cdot \left(2a_1 + (n-1) \cdot d \right) \\[1 em] S_{ 0 } &= \frac{ 0 }{2} \cdot \left( 2 \cdot \frac{ 1220 }{ 51 } + ( 0-1) \cdot 0 \right) \\[1 em] S_{ 0 } &= \frac{ 0 }{2} \cdot \left( \frac{ 2440 }{ 51 } + -1 \cdot 0 \right) \\[1 em] S_{ 0 } &= \frac{ 0 }{2} \cdot \left( \frac{ 2440 }{ 51 } + 0 \right) \\[1 em] S_{ 0 } &= \frac{ 0 }{2} \cdot \frac{ 2440 }{ 51 } \\[1 em] S_{ 0 } &= 0 \end{aligned}$$

The first few terms of this sequence are:

$$ \frac{ 1220 }{ 51 }, ~~~\frac{ 1220 }{ 51 }, ~~~\frac{ 1220 }{ 51 }, ~~~\frac{ 1220 }{ 51 }, ~~~\frac{ 1220 }{ 51 } . . . $$

Arithmetic sequences