Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
\(S_{51} = 5610\)
Step 1: Recall the basic formulas for an AP.
nth term: \(a_n = a + (n - 1)d\)
Sum of first n terms: \(S_n = \dfrac{n}{2}[2a + (n - 1)d]\)
Here, \(a\) is the first term and \(d\) is the common difference.
The second term \(a_2 = 14\) and the third term \(a_3 = 18\).
Write them using the nth-term formula:
\[a_2 = a + d = 14\]
\[a_3 = a + 2d = 18\]
Subtract the first equation from the second:
\[(a + 2d) - (a + d) = 18 - 14\]
\[a + 2d - a - d = 4\]
\[d = 4\]
Use \(a + d = 14\):
\[a + 4 = 14 \Rightarrow a = 10\]
We need \(S_{51}\). Use the sum formula:
\[S_{51} = \dfrac{51}{2}[2a + (51 - 1)d]\]
Substitute \(a = 10\) and \(d = 4\):
\[S_{51} = \dfrac{51}{2}[2 \cdot 10 + 50 \cdot 4]\]
\[= \dfrac{51}{2}[20 + 200]\]
\[= \dfrac{51}{2} \cdot 220\]
\[= 51 \cdot 110 = 5610\]
Conclusion: The sum of the first 51 terms of the AP is \(S_{51} = 5610\).