In an arithmetic progression, a_4 = 20 and a_11 = 69. What is a_1?
Answer and explanation
Correct answer: -1
Use a_n = a_1 + (n - 1)d. The two conditions become a_1 + 3d = 20 and a_1 + 10d = 69. Subtracting gives 7d = 49, so d = 7. Substitute this into the first equation: a_1 + 3(7) = 20, hence a_1 + 21 = 20 and a_1 = -1. Therefore option A is correct. Verification is straightforward: the fourth term is -1 + 3×7 = 20, while the eleventh term is -1 + 10×7 = 69. The positive alternatives would produce a fourth term greater than 20 when the common difference is 7, so they cannot satisfy both given conditions.
Frequently asked questions
What is the correct answer to this question?
-1
Why is this the correct answer?
Use a_n = a_1 + (n - 1)d. The two conditions become a_1 + 3d = 20 and a_1 + 10d = 69. Subtracting gives 7d = 49, so d = 7. Substitute this into the first equation: a_1 + 3(7) = 20, hence a_1 + 21 = 20 and a_1 = -1. Therefore option A is correct. Verification is straightforward: the fourth term is -1 + 3×7 = 20, while the eleventh term is -1 + 10×7 = 69. The positive alternatives would produce a fourth term greater than 20 when the common difference is 7, so they cannot satisfy both given conditions.
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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