SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given parameters
[tex]\begin{gathered} a_6=ar^5=\frac{17}{625} \\ a_{11}=ar^{10}=-85 \\ where\text{ a is the first term and r is the common ratio} \end{gathered}[/tex]
STEP 2: Divide the two expressions to get the common ratio
[tex]\begin{gathered} \frac{a_{11}}{a_6}=\frac{ar^{10}}{ar^5}=-\frac{85}{\frac{17}{625}} \\ a\text{ cancels a} \\ r^{10-5}=-85\div\frac{17}{625} \\ r^5=-85\times\frac{625}{17}=-5\times625=-3125 \\ r=\sqrt[5]{-3125}=-5 \\ common-ratio=-5 \end{gathered}[/tex]
STEP 3: Get the first term
[tex]\begin{gathered} From\text{ equation in step 1,} \\ ar^5=\frac{17}{625} \\ By\text{ substitution,} \\ a(-5)^5=\frac{17}{625} \\ a\times-3125=\frac{17}{625} \\ Divide\text{ both sides by -3125} \\ a=\frac{17}{625}\div-3125=\frac{17}{625}\times\frac{1}{-3125} \\ a=-\frac{17}{1953125} \end{gathered}[/tex]
STEP 4: Calculate the 14th term
[tex]\begin{gathered} a_{14}=ar^{13} \\ a_{14}=-\frac{17}{1953125}\times-5^{13} \\ a_{14}=10625 \end{gathered}[/tex]
Hence, the 14th term is 10625
