geometric series

[匿名]

2010-05-18 8:11 pm

For the geometric series 1250 , x , y , 10 then y – x =

a. –260

b. –200

c. 200

d. 250

The common ratio r of a geometric sequence is –3 and the sum of the first sixth terms S(6) is –1092. Find the first term a.

a. 3

b. 6

c. 5

d. 4

The common ratio r of the geometric series is –0.8, and the first term is 72. Find the sum to infinity of this geometric series.

a. 40

b. 36

c. 44

d. 32

A man borrowed $500,000 from a bank to buy a flat at the first day of a month. If the interest rate is 6% p.a., and the money is to be repaid in 60 equal monthly installments at the end of each month. The amount of monthly installment is

a. $10121.3

b. $9666.4

c. $7554.2

d. $8786.5

回答 (1)

Prof. Physics

2010-05-18 8:31 pm

✔ 最佳答案

1. As the sequence is a geometric sequence,

So, x/1250 = y/x = 10/y (they have a common ratio)

So, y = x^2/1250

And y^2 = 10x

(x^2/1250)^2 = 10x

x^3 = 15 625 000

x = 250

So, y = (250)^2/1250 = 50

And y - x = 50 - 250 = -200

2. Sum of the first sixth terms, a(1 - r^n)/(1 - r) = -1092

a(1 - (-3)^6)/(1 - (-3)) = -1092

First term, a = 6

3. Sum to infinity = a/(1 - r) = 72 /(1 - (-0.8)) = 40

4. Amount left at the end of first month = 500 000 X (1 + 6%/12) - A = 500 000(1.005) - A

where A is the monthly installment

Amount left at the end of second month

= [500 000(1.005) - A](1.005) - A

= 500 000(1.005)^2 - A[(1.005) + 1]

Amount left at the end of third month

= [500 000(1.005)^2 - A[(1.005) + 1]](1.005) - A

= 500 000(1.005)^3 - A[(1.005)^2 + (1.005) + 1]

...

Amount left at the end of the 60th month

= 500 000(1.005)^60 - A[(1.005)^59 + (1.005)^58 + ... + (1.005) + 1] = 0

So, A[(1.005)^59 + (1.005)^58 + ... + (1.005) + 1] = 500 000(1.005)^60

A{(1)[1 - (1.005)^60]/(1 - 1.005)} = 500 000(1.005)^60

200A[(1.005)^60 - 1] = 500 000(1.005)^60

A = $9666.4 (cor. to 1 d.p.)

參考: Prof. Physics

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