Answer
$a_n=\dfrac{3}{2}n-1$
$a_{15}=\dfrac{43}{2}$
$S_{15}=165$
Work Step by Step
The nth term is given by $a_n= a_1+(n-1) d$ ...(1)
Here, we have Common Difference $d=\dfrac{3}{2}$ and first term $a_1=\dfrac{1}{2}$
Equation (1) gives: $a_n=\dfrac{1}{2}+\dfrac{3}{2} \times (n-1)=\dfrac{3}{2}n-1$
Plugging in $n =15$, we have
$a_{15}=[\dfrac{3}{2} \times 15]-1=\dfrac{43}{2}$
We know that $S_{n}=\dfrac{n(a_1+a_{15})}{2}$
Now, $S_{15}=\dfrac{15 (\dfrac{1}{2}+\dfrac{43}{2})}{2}=165$
