What is the sum of the first 13 terms of the arithmetic progression (6, 11, 16, 21, ...)?
Answer and explanation
Correct answer: 468
The governing concept is the sum of the first n terms of an arithmetic progression: S_n = n/2 [2a + (n - 1)d]. In this sequence, a = 6, d = 11 - 6 = 5, and n = 13. Thus S_13 = 13/2 [2(6) + (13 - 1)(5)] = 13/2 [12 + 60] = 13/2 × 72 = 13 × 36 = 468. Option A is therefore correct. The same result can be checked by finding the last term, 6 + 12×5 = 66, and using average × number of terms: (6 + 66)/2 × 13 = 36×13 = 468. The other options arise from arithmetic or formula errors.
Frequently asked questions
What is the correct answer to this question?
468
Why is this the correct answer?
The governing concept is the sum of the first n terms of an arithmetic progression: S_n = n/2 [2a + (n - 1)d]. In this sequence, a = 6, d = 11 - 6 = 5, and n = 13. Thus S_13 = 13/2 [2(6) + (13 - 1)(5)] = 13/2 [12 + 60] = 13/2 × 72 = 13 × 36 = 468. Option A is therefore correct. The same result can be checked by finding the last term, 6 + 12×5 = 66, and using average × number of terms: (6 + 66)/2 × 13 = 36×13 = 468. The other options arise from arithmetic or formula errors.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Arithmetic Progression.
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