Question

Find the 10th term of the series $-\frac{1}{2},\frac{1}{2},\frac{3}{2}$

Answer 1

Given,

$a=-\frac{1}{2}$

$d=\frac{1}{2}-\left(\frac{1}{2}\right)=1$

Formula:

$T_n=a+(n-1)d$

$T_{10}=-\frac{1}{2}+(10-1)1$

$=-\frac{1}{2}+9$

$=\frac{17}{2}$