Question

$25th$ term of the $A.P:5,\frac{5}{2},0,-\frac{5}{2},...$

Answer 1

$5,\frac{5}{2},0,-\frac{5}{2},..$ are in A.P

where the first term$=a=\frac{5}{2}$ and the common difference

$d=\frac{5}{2}-5=\frac{5-10}{2}=\frac{-5}{2}$

$n^{th}$ term$=t_n=a+(n-1)d$

$t_{25}=\frac{5}{2}+(25-1)\frac{-5}{2}$

$=\frac{5}{2}-\frac{24\times 5}{2}$

$=\frac{5-120}{2}=\frac{-115}{2}$

$\therefore {25}^{th}$ term of an A.P is $\frac{-115}{2}$