If \(a_n=6\cdot2^{n-1}\) what first three terms does it give?
Answer and explanation
Correct answer: (6, 12, 24)
Given \(a_n=6\cdot2^{n-1}\), for \(n=1\), \(a_1=6\cdot2^0=6\); for \(n=2\), \(a_2=6\cdot2^1=12\); and for \(n=3\), \(a_3=6\cdot2^2=24\). Therefore, the first three terms are \((6, 12, 24)\). Option B misses the initial coefficient 6. Exam tip: always substitute \(n=1\) first to check the first term.
Frequently asked questions
What is the correct answer to this question?
(6, 12, 24)
Why is this the correct answer?
Given \(a_n=6\cdot2^{n-1}\), for \(n=1\), \(a_1=6\cdot2^0=6\); for \(n=2\), \(a_2=6\cdot2^1=12\); and for \(n=3\), \(a_3=6\cdot2^2=24\). Therefore, the first three terms are \((6, 12, 24)\). Option B misses the initial coefficient 6. Exam tip: always substitute \(n=1\) first to check the first term.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Geometric Progression.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
