sequence Questions (maths)

2008-09-05 1:51 am
1) the arithmertic mean and the geometric mean of two positive integers are 5 and 4 respectively, Find the two integers.

2) in a geometric sequence, T(2) x T(5) = 243 and T(4) = 27. Find the general term, T(n),of the sequence.

3) in the begining of a certain year, country A and country B have populations 200000 and 180000 respectively, if the populations of country A and country B are increased by 0.5% and 2% each year respectively, how many year will it take for country B to have a higher population than country A?

4) if x-4, x+4, 4x+4 are in geometric sequence, find the possible values of x.


thank you so much , hope anyone can help me on today ! thanks!

回答 (1)

2008-09-05 3:15 am
✔ 最佳答案
1)
Let p and q be the two integers.

Arithmetic mean: (p+q)/2 = 5
Hence, p = 10 - q ..... [1]

Geometric mean: √(pq) = 4
Hence, pq = 16 ..... [2]

Put [1] into [2]:
(10 - q)q = 16
q2 - 10q + 16 = 0
(q - 2)(q + 8) = 0
q = 2 ooro q = 8

When q = 2, p = 8
When q = 8, p = 2

Ans: The two integers are 2 and 8.

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2)
Let a and r be the 1st term and the common ratio respectively.

T(2) x T(5) = 243
ar x ar4 = 243
a2r5 = 243 ..... [1]

T(4) = 27
ar3 = 27 ..... [2]
a2r6 = 729 ..... [3]

[3]/[1]
a2r6/(a2r5) = 729/243
r = 3

Put r = 3 into [2]:
a(3)3 = 27
a = 1

Hence, T(n) = arn-1
Hence, T(n) = 3n-1

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3)
Let n be the required number of years.

180000 (1+2%)n > 200000(1+0.5%)n
9(1.02)n > 10(1.005)n
(1.02)n/(1.005)n > 10/9
(1.02/1.005)n > (10/9)
log(1.02/1.005)n > log(10/9)
n log(1.02/1.005) > log(10/9)
n(log1.02-log1.005) > log10-log9
n > (log10-log9)/(log1.02-log1.005)
n > 7.11

Ans: It takes 8 years.

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4)
The geometric mean:
√[(x - 4)(4x + 4)] = x + 4
[√(4x2 - 12x - 16)]2 = (x + 4)2
4x2 - 12x - 16 = x2 + 8x + 16
3x2 - 20x - 32 = 0
(3x + 4)(x - 8) = 0
x = -4/3 ooro x = 8
=


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