sum of an arithmetic progression

Let the first term of the AP be $a$ and the common difference be $d$ . As $n_{th}$ term of an AP is defined as $a+(n-1)d$ therefore

$a+2d=55$

$a+15d=315$

If we subtract our first equation from our second equation we get

$a+15d-a-2d=315-55$

$13d=260$

$d=20$

If we put this in our first equation

$a+2(20)=55$

$a=55-2(20)=55-40=15$

Now as sum of $n$ terms of an AP is $\frac{n}{2}[2a+(n-1)d]$ (see why? )

$Sum=\frac{18}{2}[2(15)+(18-1)20]=9[30+(17)20]=9[30+340]=9[370]=\boxed{3330}$

According to $a_n = a_1 + (n - 1)*d$ , we can get {d = 20 & a1 = 15} by $a3 = a1 + (3 - 1)*d$ and $a16 = a1 + (16 - 1)*d$ So the sum of this terms = $(\frac{18(2*15 + 17 * 20)}{2}$ ) = 3330