Which geometric progression has first term 4 and common ratio 3?
Answer and explanation
Correct answer: \((4,12,36,108,\ldots)\)
A geometric progression with first term 4 and common ratio 3 must begin with 4, and every following term must be obtained by multiplying the preceding term by 3. Option A gives \(4,4\times3=12,12\times3=36,36\times3=108\), so it satisfies both conditions. Option B begins with 3, option C begins correctly but adds 4 rather than multiplying by 3, and option D has ratio 3 but begins with 12 instead of 4. Therefore, option A is the only sequence meeting the first-term and common-ratio requirements simultaneously.
Frequently asked questions
What is the correct answer to this question?
\((4,12,36,108,\ldots)\)
Why is this the correct answer?
A geometric progression with first term 4 and common ratio 3 must begin with 4, and every following term must be obtained by multiplying the preceding term by 3. Option A gives \(4,4\times3=12,12\times3=36,36\times3=108\), so it satisfies both conditions. Option B begins with 3, option C begins correctly but adds 4 rather than multiplying by 3, and option D has ratio 3 but begins with 12 instead of 4. Therefore, option A is the only sequence meeting the first-term and common-ratio requirements simultaneously.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Geometric Progression.
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