Question

Find the common difference and $99th$ term of the arithmetic progression:
$7\frac{3}{4},9\frac{1}{2},11\frac{1}{4},............$

Answer 1

$7\frac{3}{4},9\frac{1}{2},11\frac{1}{4},...............$

$=\frac{31}{4},\frac{19}{2},\frac{45}{4},..............$

$a=\frac{31}{4},d=\frac{19}{2}-\frac{31}{4}=\frac{45}{4}-\frac{19}{2}=\frac{7}{4}$

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

$T_{99}=\frac{31}{4}+(99-1)\frac{7}{4}$

$=\frac{31}{4}+\left(98\right)\frac{7}{4}$

$=\frac{31}{4}+\frac{686}{4}$

$=\frac{717}{4}$