Question

For an arithmetic progression: $5,\frac{11}{2},6,\frac{13}{2},\dots$ then $T_{40}-T_{20}=?$

Answer 1

$d=\frac{11}{2}-5=\frac{1}{2}$

Formula,

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

$T_{40}-T_{20}$

$=a+39d-a-19d$

$=20d$

$=20\times \frac{1}{2}$

$=10$