Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Step-by-Step Solution
Key Concept: Use the definition of an arithmetic progression to determine the first term \(a\) and common difference \(d\) from the given consecutive terms, then apply the sum formula \(S_n = \frac{n}{2}[2a+(n-1)d]\) (or \(S_n = \frac{n}{2}(a+l)\)).
1. Let the first term be \(a\) and the common difference be \(d\).\
2. From the given information:\
\[a_2 = a + d = 14\]\
\[a_3 = a + 2d = 18\]\
3. Subtract the two equations to find \(d\):\
\[(a+2d)-(a+d) = 18-14 \Rightarrow d = 4\].\
4. Substitute \(d = 4\) into \(a + d = 14\) to get \(a\):\
\[a = 14 - d = 14 - 4 = 10\].\
5. Number of terms required, \(n = 51\).\
6. Compute the last term \(l\) (or \(a_{51}\)) using \(l = a + (n-1)d\):\
\[l = 10 + (51-1)\times 4 = 10 + 50\times 4 = 10 + 200 = 210\].\
7. Use the sum formula for an AP:\
\[S_{51} = \frac{n}{2}(a + l) = \frac{51}{2}(10 + 210)\].\
8. Simplify:\
\[S_{51} = \frac{51}{2}\times 220 = 51 \times 110 = 5610\].\
9. Hence, the sum of the first 51 terms is \(5610\).
Correct Answer: 5610