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A sequence contains terms A_1
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16 May 2024, 01:26
16 May 2024, 01:26
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71% (01:27) correct
28% (02:31) wrong based on 14 sessions
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A sequence contains terms \(A_1, A_2,A_3,..........A_n\) such that \(A_1=\frac{1}{2}\) and general term is given as \(A_n=\frac{n}{(n+1)} * A_{(n-1)}\)
Quantity A |
Quantity B |
Value of \(k\), if \(A_k=\frac{1}{15}\) |
\(14\) |
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A sequence contains terms A_1
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16 May 2024, 08:07
16 May 2024, 08:07
1
Calculate the first few terms of the sequence after \(A_1\) to identify a pattern.
\(A_2 = \frac{2}{(2+1)} * A_{(2-1)} = \frac{2}{3} * A_1 = \frac{2}{3}*\frac{1}{2} = \frac{1}{3}\)
\(A_3 = \frac{3}{(3+1)}* A_2 = \frac{3}{4} *\frac{1}{3} = \frac{1}{4}\)
We can see that \(A_n\) can also be given as \(A_n = \frac{1}{n+1}\).
Given that \(A_k=\frac{1}{15}\),
\(15 = k+1\)
\(k = 14\)
Thus, Quantities A and B are equal.
\(A_2 = \frac{2}{(2+1)} * A_{(2-1)} = \frac{2}{3} * A_1 = \frac{2}{3}*\frac{1}{2} = \frac{1}{3}\)
\(A_3 = \frac{3}{(3+1)}* A_2 = \frac{3}{4} *\frac{1}{3} = \frac{1}{4}\)
We can see that \(A_n\) can also be given as \(A_n = \frac{1}{n+1}\).
Given that \(A_k=\frac{1}{15}\),
\(15 = k+1\)
\(k = 14\)
Thus, Quantities A and B are equal.
gmatclubot
A sequence contains terms A_1 [#permalink]
16 May 2024, 08:07
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