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705-805 (Hard)|   Sequences|      
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nick1816
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A sequence is given by the rule a_{n+2}= 2/a_{n+1}+a_n, n is a positi 31 Oct 2019, 09:05
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D
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85% (hard)

Question Stats:

55% (03:33) correct 45% (02:58) wrong based on 160 sessions

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A sequence is given by the rule

\(a_{n+2}=\frac{2}{a_{n+1}} +a_n\), where n>0. Also, \(a_1\)=1 and \(a_2\)=2.

Find the value of \(a_{100}*a_{101}\)?

A. 80
B. 120
C. 160
D. 200
E. 240
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D
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Peddi
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A sequence is given by the rule a_{n+2}= 2/a_{n+1}+a_n, n is a positi 31 Oct 2019, 09:43
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A101=(2/A100)+A99. ..>A100*A101=2+A99.A100=2+2+A98.A99.....the sequence follows the final answersA100.A101=2*100=200

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Anjali_1453
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A sequence is given by the rule a_{n+2}= 2/a_{n+1}+a_n, n is a positi 25 May 2024, 22:32
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Can somebody explain this question with solution

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Oppenheimer1945
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A sequence is given by the rule a_{n+2}= 2/a_{n+1}+a_n, n is a positi 26 May 2024, 08:58
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Anjali_1453
Can somebody explain this question with solution

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­a_{n+2} a_{n+1}-a_{n+1}a_{n}=2

put n=1,2,3,...99:
a3a2-a2a1=2
a4a3-a3a2=2
..
.
a101a100-a100a99=2

add it up:
a101a100-a2a1=2(99)
=> a101a100=198+2=200
 
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hienthusiast
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A sequence is given by the rule a_{n+2}= 2/a_{n+1}+a_n, n is a positi 18 Sep 2024, 23:50
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\(a_3\) = \(\frac{2}{a_2}\) + \(a_{1}\) => \(a_{3}\) = \(\frac{2 + a_1 * a_2}{a_2}\) => \(a_2\)*\(a_3\) = 2 + \(a_1\)*\(a_2\)
\(a_4\) = \(\frac{2}{a_3}\) + \(a_{2}\) => \(a_{4}\) = \(\frac{2 + a_2 * a_3}{a_3}\) => \(a_3\)*\(a_4\) = 2 + \(a_2\)*\(a_3\) = 2*2 + \(a_1\)*\(a_2\)
\(a_5\) = \(\frac{2}{a_4}\) + \(a_{3}\) => \(a_{5}\) = \(\frac{2 + a_3 * a_4}{a_4}\) => \(a_4\)*\(a_5\) = 2 + \(a_3\)*\(a_4\) = 2*3 + \(a_1\)*\(a_2\)

New rule of the sequence: \(a_{n}\)*\(a_{n+1}\) = 2*(n-1) + \(a_1\)*\(a_2\)

Therefore, \(a_{100}\)*\(a_{101}\) = 2*99 + \(a_1\)*\(a_2\) = 198 + 2 = 200

The answer is D
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Vivek1707
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A sequence is given by the rule a_{n+2}= 2/a_{n+1}+a_n, n is a positi 31 Jan 2026, 11:47
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Key is to identify the pattern

a1 = 1
a2 = 2
a3 = 2/2+1 = 2
a4 = 2/2+2 = 3
a5 = 2/3+2 i.e 8/3
a6 = 2/(8/3) +3 i.e 15/4

If you try making pairs you'll notice that the result is consecutive even numbers i.e
a1 x a2 = 2
a2 x a3 = 4
a3 x a4 = 6
a4 x a5 = 8

Hence
a100 x a101 = 200 i.e D
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A sequence is given by the rule a_{n+2}= 2/a_{n+1}+a_n, n is a positi 31 Jan 2026, 12:11
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Can you explain how you extrapolated from the patern of a1xa2 to a4xa5 to see that a100xa101 is 200?
Vivek1707
Key is to identify the pattern

a1 = 1
a2 = 2
a3 = 2/2+1 = 2
a4 = 2/2+2 = 3
a5 = 2/3+2 i.e 8/3
a6 = 2/(8/3) +3 i.e 15/4

If you try making pairs you'll notice that the result is consecutive even numbers i.e
a1 x a2 = 2
a2 x a3 = 4
a3 x a4 = 6
a4 x a5 = 8

Hence
a100 x a101 = 200 i.e D
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Vivek1707
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A sequence is given by the rule a_{n+2}= 2/a_{n+1}+a_n, n is a positi 31 Jan 2026, 12:19
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Notice that the even numbers that you get by multiplying each pair of terms are just twice nth value of the first term in the pair
i.e

a1 x a2 = 2 -----> 1 x 2 = 2
a2 x a3 = 4 ----> 2 x 2 = 4
a3 x a4 = 6 ----> 3 x 2 = 6
a4 x a5 = 8 ----> 4 x 2 = 8

a100x a101 ----> 100 x 2 = 200

Hope this helps
roshaun25
Can you explain how you extrapolated from the patern of a1xa2 to a4xa5 to see that a100xa101 is 200?

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