Consider the sequence a_(n)=(3n-4)/(2n-1) Find a_(1) and a_(2) a_(1)=(2)/(3) and a_(2)=-1 a_(1)=-1 and a_(2)=(3)/(2) a_(1)=-1 and a_(2)=(2)/(3) a_(1)=1 and a_(2)=(2)/(3) a_(1)=1 and a_(2)=-(3)/(2)

## Step 1: Calculate $a_1$<br />### Substitute $n=1$ into the formula $a_{n}=\frac{3n-4}{2n-1}$. $a_1 = \frac{3(1)-4}{2(1)-1} = \frac{3-4}{2-1} = \frac{-1}{1} = -1$<br /><br />## Step 2: Calculate $a_2$<br />### Substitute $n=2$ into the formula $a_{n}=\frac{3n-4}{2n-1}$. $a_2 = \frac{3(2)-4}{2(2)-1} = \frac{6-4}{4-1} = \frac{2}{3}$