If \(a_n=3\cdot2^{n-1}\), what first three terms does it give?
Answer and explanation
Correct answer: (3, 6, 12)
Given \(a_n=3\cdot2^{n-1}\): for \(n=1\), \(a_1=3\cdot2^0=3\); for \(n=2\), \(a_2=3\cdot2^1=6\); and for \(n=3\), \(a_3=3\cdot2^2=12\). Hence, the first three terms are \((3, 6, 12)\). The sequence \((3, 9, 27)\) has common ratio 3, whereas this sequence has common ratio 2. Exam tip: substitute \(n=1\) first and check the exponent \(n-1\) carefully.
Frequently asked questions
What is the correct answer to this question?
(3, 6, 12)
Why is this the correct answer?
Given \(a_n=3\cdot2^{n-1}\): for \(n=1\), \(a_1=3\cdot2^0=3\); for \(n=2\), \(a_2=3\cdot2^1=6\); and for \(n=3\), \(a_3=3\cdot2^2=12\). Hence, the first three terms are \((3, 6, 12)\). The sequence \((3, 9, 27)\) has common ratio 3, whereas this sequence has common ratio 2. Exam tip: substitute \(n=1\) first and check the exponent \(n-1\) carefully.
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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