The sum of the squares of the fifth and the eleventh term of an AP is $3$ and the product of the second and fourteenth term is equal to $x$. Find the product of the first and fifteenth term of an AP.
- $\dfrac{(58x-39)}{45} \\$
- $\dfrac{(98x-39)}{72} \\$
- $\dfrac{(116x-39)}{90} \\$
- $\dfrac{(98x-39)}{90}$