If the 5th and 12th terms of an A.P. are 30 and 65 respectively,

Solution:

We have;

$a_{5}=30$

$\Rightarrow a+(5-1) d=30$

$\Rightarrow a+4 d=30 \quad \ldots(\mathrm{i})$

Also, $a_{12}=65$

$\Rightarrow a+(12-1) d=65$

$\Rightarrow a+11 d=65 \quad \ldots \ldots($ ii $)$

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

$7 d=35$

$\Rightarrow d=5$

Putting the value of $d$ in (i), we get:

$a+4 \times 5=30$

$\Rightarrow a=10$

$\therefore S_{20}=\frac{20}{2}[2 \times 10+(20-1) \times 5]$

$\Rightarrow S_{20}=10[2 \times 10+(20-1) \times 5]$

$\Rightarrow S_{20}=1150$