Answer
$234$
Work Step by Step
$a_{n+1}-a_n=\frac{1}{3}(n+1)+4-\left(\frac{1}{3}n+4\right)=\frac{1}{3}$
Since the difference between consecutive terms is constant, the given sequence is arithmetic.
The formula for the finite sum of the arithmetic sequence is:
$S_{n}=\frac{n(a_{1}+a_{n})}{2}$
$a_{n}=\frac{1}{3}n+4$
$a_{1}=\frac{1}{3}(1)+4=\frac{13}{3}$
.
$a_{27}=\frac{1}{3}(27)+4=13$
$S_{27}=\frac{27(a_{1}+a_{27})}{2}=\frac{27\left(\frac{13}{3}+13\right)}{2}=234$
