If \(a_n=4\cdot3^{n-1}\) what first three terms does it give?
Answer and explanation
Correct answer: (4, 12, 36)
For \(n=1\), \(a_1=4\cdot3^0=4\); for \(n=2\), \(a_2=4\cdot3^1=12\); and for \(n=3\), \(a_3=4\cdot3^2=36\). Therefore, the first three terms are \((4, 12, 36)\). In \((3, 9, 27)\), the first term is 3, but the given formula gives \(a_1=4\). Exam tip: substitute \(n=1\) first to verify the initial term.
Frequently asked questions
What is the correct answer to this question?
(4, 12, 36)
Why is this the correct answer?
For \(n=1\), \(a_1=4\cdot3^0=4\); for \(n=2\), \(a_2=4\cdot3^1=12\); and for \(n=3\), \(a_3=4\cdot3^2=36\). Therefore, the first three terms are \((4, 12, 36)\). In \((3, 9, 27)\), the first term is 3, but the given formula gives \(a_1=4\). Exam tip: substitute \(n=1\) first to verify the initial term.
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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