Answer:
[tex]\frac{8,671}{6}[/tex]
Explanation:
Here, we want to get the sum of the 58 terms in series
Mathematically, we have the formula to use as:
[tex]S_n\text{ = }\frac{n}{2}(a\text{ + L)}[/tex]
where a is the first term and L is the last term
The first term is when n is 1
We have this calculated as:
[tex]\text{ a}_{}\text{ = }\frac{5}{6}+\frac{1}{3}\text{ = }\frac{5+2\text{ }}{6}\text{ = }\frac{7}{6}[/tex]
The last term is the 58th term which is:
[tex]\text{ a}_{58}\text{ = }\frac{290}{6}\text{ + }\frac{1}{3}\text{ = }\frac{292}{6}[/tex]
We finally substitute these values into the initial equation
Thus, we have it that:
[tex]S_{58}\text{ = }\frac{58}{2}(\frac{292}{6}+\frac{7}{6})\text{ = 29(}\frac{299}{6})\text{ = }\frac{8671}{6}[/tex]
