Let a be the first term and d be the common difference of the AP. Then,
d = a3 - a2 = 18 - 14 = 4
Now,
a2 = 14 (Given)
\(\Rightarrow\) a + d = 14 [an = a+(n - 1)d]
\(\Rightarrow\) a + 4 = 14
\(\Rightarrow\) a = 14 - 4 = 10
Using the formula, \(S_n=\frac{n}{2}[2a+(n-1)d],\) we get
\(S_{51}=\frac{51}{2}[2\times10+(51-1)\times4]\)
= \(\frac{51}{2}(20+200)\)
= \(\frac{51}{2}\times220\)
= 5610
Hence, the required sum is 5610.
