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C
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Difficulty:
95%
(hard)
Question Stats:
43% (02:35) correct
57%
(02:32)
wrong
based on 157
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There is a sequence A(n) where n is a positive integer such that A(n+1) = 10 + 0.5A(n). Which of the following is closest to A(1,000)?
A. 15
B. 18
C. 20
D. 25
E. 50
A. 15
B. 18
C. 20
D. 25
E. 50
15 Jan 2017, 01:00
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hwang327
Hi
This is combination of geometric and arithmetic sequences.
n>0, min n = 1.
\(a_2 = 10 + \frac{1}{2}a_1\)
\(a_3 = 10 + \frac{1}{2}a_2 = 10 + \frac{1}{2}(10 + \frac{1}{2}a_1) = 10 + 5 + \frac{1}{4}a_1\)
\(a_4 = 10 + \frac{1}{2}a_3 = 10 + \frac{1}{2}(10 + 5 + \frac{1}{4}a_1) = 10 + 5 + \frac{5}{2} + \frac{1}{8}a_1\)
\(a_5 = 10 + \frac{1}{2}a_4 = 10 + 5 + \frac{5}{2} + \frac{5}{4} + \frac{1}{16}a_1\)
...
\(a_n = 10 + 5 + \frac{5}{2} + \frac{5}{4} + ... + \frac{5}{2^{n-3}} + \frac{1}{2}^{n-1}a_1\)
When n=1000 our fraction \(\frac{1}{2^{999}}\) is close to 0.
\(5 + \frac{5}{2} + \frac{5}{4} + ... + \frac{5}{2^{997}}\) We can apply logic of infinite geometric series where |r|<1 because our n is quite big (1000).
\(S = \frac{5}{1-1/2} = 5*2 = 10\)
\(a_{1000} ≈ 10 + 10 + 0 = 20\)
Answer C
Hope this helps
Regards
General Discussion
15 Jan 2017, 04:49
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hwang327
Hi,
A point before the solution..
The Q is flawed in that there is no value of \(A_1\) given.
Solution..
\(A_1=10, A_2=10+0.5*10=10+5, A_3=10+5+0.5*5=10+5+0.25=10+10/2+10/4+.....\)
So 1000 can be taken as infinite series..
Ans =\(\frac{a}{(1-r)}=10/(1-1/2)=10/(1/2)=20\)
