Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Given that,
Second term, \(a_2\) = 14
Third term, \(a_3\) = 18
Common difference, d
= \(a_3 - a_2\) = 18-14 = 4
\(a_2\) = a + d
14 = a + 4
a = 10 = First term

Sum of n terms;

\(S_n = {{n} \over {2}} [2a + (n – 1)d]\)
\(S_{51} = {{51} \over {2}} [2×10 (51-1) 4]\)
= \({{51} \over {2}}[2 + (20)4\)]
= 51×\({{220} \over {2}}\)
= 51×110
= 5610