Answer
The sequence if arithimetic.
$d=-\dfrac{3}{4}$
$S_{50}=-\dfrac{2225}{4}$
Work Step by Step
$\bf\text{RECALL:}$
$\bf\text{(1) Arithmetic Sequence }$
A sequence is arithmetic if there exists a common difference $d$ among consecutive terms.
$d=a_n-a_{n-1}$
The sum of the first $n$ terms of an arithmetic sequence is given by the formulas:
$S_n=\frac{n}{2}(a_1 +a_n)$
or
$S_n=\frac{n}{2}\left(2 a_1 + (n-1)d\right)$
$\bf\text{(2) Geometric Sequence }$
A sequence is geometric if there exists a common ratio $r$ among consecutive terms.
$r=\dfrac{a_n}{a_{n-1}}$
The sum of the first $n$ terms of a geometric sequence is given by the formula:
$S_{n}=a_1 \cdot \dfrac{1-r^n}{1-r}$
In the formulas listed above,
$d$ = common difference
$r$ = common ratio
$a_1$ = first term
$a_n$ = nth term
$n$ = number of terms
$\bf\text{List the first few terms of the sequence.}$
$\bf\text{Identify the sequence as arithmetic or geometric.}$
Substitute $1, 2, 3$ for $n$ to list the first three terms:
$a_1 =8-\frac{3}{4}(1)=8-\frac{3}{4} = \frac{32}{4} - \frac{3}{4} =\frac{29}{4}$
$a_2 = 8-\frac{3}{4}(2)=8-\frac{6}{4} = \frac{32}{4} - \frac{6}{4} =\frac{26}{4}=\frac{13}{2}$
$a_3 = 8-\frac{3}{4}(3)=8-\frac{9}{4} = \frac{32}{4} - \frac{9}{4} =\frac{23}{4}$
Notice that the values decrease by $\frac{3}{4}$.
Thus, the sequence is arithmetic with $d=-\frac{3}{4}$.
$\bf\text{Find the sum of the first 50 terms}:$
With $a_1=\dfrac{29}{4}$ and $d=-\frac{3}{4}$, solve for the sum of the first 50 terms using the formula in (1) above to obtain:
$S_n = \dfrac{n}{2}\left(2a_1+(n-1)d)\right)
\\S_{50} = \dfrac{50}{2}\left(2\cdot(\frac{29}{4}) + (-\frac{3}{4})(50-1)\right)
\\S_{50} = 25(\frac{58}{4}+(-\frac{3}{4}) \cdot 49)
\\S_{50} = 25(\frac{58}{4} -\frac{147}{4})
\\S_{50}=25(-\frac{89}{4})
\\S_{50}=-\dfrac{2225}{4}$
