Answer
$d=0.5$
$a_1=\dfrac{1}{3}$
$a_2=\dfrac{5}{6}$
$a_3=\dfrac{4}{3}$
$a_4=\dfrac{11}{6}$
$a_5=\dfrac{7}{3}$
Work Step by Step
We have an arithmetic sequence in which $a_{17}=\dfrac{25}{3}$ and $a_{32}=\dfrac{95}{6}.$ We have to find first five terms and $a_1,$ $d.$ In order to find $a_{1}$ and $d$ we will use the fact that any term of arithmetic sequence is expressed through the formula $a_{n}=a_1+d(n-1)$
Hence,
$a_{17}=a_1+d\cdot16$
$a_{32}=a_1+d\cdot31$
$a_{17}-a_{32}=a_1+d\cdot16-a_1-d\cdot31=d(16-31)=-15d$
$\dfrac{25}{3}-\dfrac{95}{6}=-15d$
$-\dfrac{45}{6}=-15d$
And $d=0.5$
Now we can use any equation to find $a_1.$
$\dfrac{25}{3}=a_1+0.5\cdot16$
$\dfrac{25}{3}-8=a_1$
$a_1=\dfrac{1}{3}$
The first five terms will be
$a_1=\dfrac{1}{3}$
$a_2=a_1+d=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}$
$a_3=a_1+2d=\dfrac{1}{3}+1=\dfrac{4}{3}$
$a_4=a_1+3d=\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{11}{6}$
$a_5=a_1+4d=\dfrac{1}{3}+2=\dfrac{7}{3}$
