Maths!!請幫幫忙!

2008-09-08 3:00 am

2. In a sequence, the sum of first n terms is S(n )=2n^2+3n for all values of n. Find the 100th term.
4. Find the value(s) of x such that x+9, x-6 and 4 are in geometric sequence.
8. The third term of an arithmetic sequence is 25 and the tenth term is -3. Find the first term and the common difference.

更新1:

可否詳細解釋第4題為何(x-6)/(x+9) = 4/(x-6)?

回答 (1)

Uncle Michael

2008-09-08 3:25 am

✔ 最佳答案

2.
Sum of the first 100 terms, S(100)
= 2(100)2 + 3(100)
= 20300

Sum of the first 99 terms, S(99)
= 2(99)2 + 3(99)
= 19899

The 100th term
= S(100) - S(99)
= 20300 - 19899
= 401

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4.
Consider the common ratio:
(x-6)/(x+9) = 4/(x-6)
(x-6)2 = 4(x+9)
x2-12x+36 = 4x+36
x2-16 = 0
x(x-16) = 0
x = 0 ooro x = 16

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8.
Let a and d be the first term and the common difference.

The 3th term:
a + 2d = 25 ...... (1)

The 1oth term:
a + 9d = -3 ...... (2)

(2)-(1):
7d = -28
d = -4

Put d=-4 into (1):
a + 2(-4) = 25
a = 33

Ans: The first term is 33, and the common difference is -4.
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2008-09-08 01:23:34 補充：
In Q.4:

Consider the first two terms, common ratio = (x-6)/(x+9)

Consider the last two terms, common ratio = 4/(x-6)

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