Question

Determine the $AP$ whose third term is $16$ and ${17}^{th}$ term exceed the $5^{th}$ term by $12$.

Answer 1

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

$T_3=a+(3-1)d=a+2d$

$a+2d=16$...(1)

$T_{17}-T_5=a+16d-(a+4d)$

$a+16d-a-4d=12$

$12d=12$

$\therefore d=1$

$a+2d=16$ From (1)

$a=16-2\left(1\right)$

$\therefore a=14$

Therefore the series is:

$14,15,16,17,.............$