Find the sum of the first (22) terms of the AP (160,151,142,\ldots).

Correct answer: (1441)

The direct answer is option D: 1441. The sequence 160, 151, 142, … is an arithmetic progression because each term decreases by 9, so a=160, d=151−160=−9, and n=22. Use \\(S_n=\\frac{n}{2}[2a+(n-1)d]\\). First, n−1=21. Next, (n−1)d=21×(−9)=−189. Also, 2a=2×160=320. Therefore the bracket is 320−189=131. Now \\(S_{22}=\\frac{22}{2}×131=11×131=1441\\). Thus option D is correct. Option A, 1397, is wrong because it does not follow from the correct negative difference calculation. Option B, 1419, is also incorrect and may arise from a small subtraction error. Option C, 1463, is wrong because it is too large; using d=+9 instead of d=−9 would incorrectly increase the sum. Option D is correct because a decreasing AP must use a negative common difference. A second check uses the last term: \\(a_{22}=160+21(−9)=−29\\). The average of the first and last terms is \\((160−29)/2=65.5\\), and 22×65.5=1441. Remember: when terms decrease, d is negative, even though the terms themselves may begin with a large positive number.