Answer
$S_6=441$
Work Step by Step
The given geometric sequence has:
$a_3=28
\\a_6=224$
Note that using the third term as starting or reference point, the sixth term of the sequence can be computed by multiplying the common ratio $r$ three times to the third term.
Thus,
$a_6=a_3 \cdot r \cdot r \cdot r
\\a_6 = a_3 \cdot r^3$
Substitute the values of $a_3$ and $a_6$ into the equation above to obtain:
$a_6 = a_3 \cdot r^3
\\224 = 28 \cdot r^3
\\\dfrac{224}{28} = \dfrac{28r^3}{28}
\\8 = r^3
\\2^3 = r^3$
Take the cube root of both sides to obtain:
$2=r$
RECALL:
The partial sum $S_n$ (sum of the first $n$ terms) of a geometric sequence is given by the formula:
$S_n=a\left(\dfrac{1-r^n}{1-r}\right), r\ne 1$
where
$a$ = first term
$r$ = common ratio
As of now, only the value of $r$ is known.
We need to find the value of $a$.
Note that to find the value of the third term, the common ratio $r$ has to be multiplied twice to the first term. This means that to find the value of the first term, you can divide the common ratio twice (or $r^2$) to the third term.
Thus,
$a = a_3 \div r \div r
\\a = a_3 \div r^2$
Substitute the values of $a_3$ and $r$ to obtain:
$a=28 \div (2^2)
\\a = 28 \div 4
\\a=7$
Now that both $a$ and $r$ are known, the sum of the first 6 terms can be computed using the formula above to obtain:
$\require{cancel}
S_6 =7\left(\dfrac{1-2^6}{1-2}\right)
\\S_6=7\left(\dfrac{1-64}{-1}\right)
\\S_6=7\left(\dfrac{-63}{-1}\right)
\\S_6=7\cdot 63
\\S_6=441$
