Find the sum of first 22 terms of an AP in which \(d = 7\) and the 22nd term is 149.
Sum = \(1661\)
Given: Common difference \(d = 7\), 22nd term \(a_{22} = 149\). We need \(S_{22}\).
Step 1: Recall the formulas.
nth term of an AP: \(a_n = a + (n - 1)d\)
Sum of first n terms: \(S_n = \dfrac{n}{2}(a + l)\), where \(l = a_n\) is the nth (last) term.
Step 2: Find the first term \(a\).
For the 22nd term:
\[a_{22} = a + 21d\]
Substitute \(a_{22} = 149\) and \(d = 7\):
\[149 = a + 21 \cdot 7\]
\[149 = a + 147\]
\[a = 149 - 147 = 2\]
So, the first term is \(a = 2\).
Step 3: Use the sum formula.
We know \(n = 22\), \(a = 2\), and the last term \(l = a_{22} = 149\).
\[S_{22} = \dfrac{22}{2}(a + l) = 11(2 + 149)\]
\[S_{22} = 11 \cdot 151 = 1661\]
Conclusion: The sum of the first 22 terms of the AP is \(\boxed{1661}\).