Question

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

Answer 1

Given that, \(a_2 = 14\) and \(a_3 = 18\)
\(d = a_3 − a_2 = 18 − 14 = 4\)
\(a_2 = a + d\)
\(14 = a + 4\)
\(a = 10\)
\(Sn = \frac n2[2a + (n-1)d]\)

\(S_{51} = \frac {51}{2} [2 \times 10 + (50-1)4]\)

\(S_{51}= \frac {51}{2} [20 + (50)4]\)

\(S_{51}= \frac {51 \times 220}{2}\)
\(S_{51}= 51 \times 110\)
\(S_{51} = 5610\)