The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2.

Solution:

Given:

$a_{3}=7, a_{7}-3 a_{3}=2$

We have:

$a_{3}=7$

$\Rightarrow a+(3-1) d=7$

$\Rightarrow a+2 d=7$    ...(i)

Also, $a_{7}-3 a_{3}=2$

$\Rightarrow a_{7}-21=2$ (Given)

$\Rightarrow a+(7-1) d=23$

$\Rightarrow a+6 d=23$    ...(ii)

From (i) and (ii), we get:

$4 d=16$

$\Rightarrow d=4$

Putting the value in (i), we get:

$a+2(4)=7$

$\Rightarrow a=-1$

$\therefore S_{20}=\frac{20}{2}[2(-1)+(20-1)(4)]$

$\Rightarrow S_{20}=10[-2+76]$

$\Rightarrow S_{20}=10[74]=740$

$\therefore a=-1, d=4, S_{20}=740$