## Algebra 2 (1st Edition)

$a_n=\dfrac{3}{2}n-1$ $a_{15}=\dfrac{43}{2}$ $S_{15}=165$
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, $S_{15}=\dfrac{15 (\dfrac{1}{2}+\dfrac{43}{2})}{2}=165$