# How do you find an for the arithmetic series S_16=856 and a_1=76?

Well the sum of an arithmetic series is given by

${S}_{n} = \frac{n}{2} \cdot \left({a}_{1} + {a}_{n}\right)$

Now hence ${S}_{16} = 856$ and ${a}_{1} = 76$

we have that

$856 = \frac{16}{2} \cdot \left(76 + {a}_{16}\right) \implies {a}_{16} = 107 - 76 \implies {a}_{16} = 31$

and a_n=a_1+(n-1)*r=>a_16=a_1+15*r=> r=(a_16-a_1)/15=>r=(31-76)/15=>r=-3

So it is an arithemtic progression with first term ${a}_{1} = 76$
and ratio $r = - 3$