超難 arithemetic sequence 問題
2008-01-01 10:45 pm
1. If the ratio of the sum of the first n terms of two arithmetic sequences is (2+3n):(3+2n), find the ratio of their seventh terms.
回答 (3)
2008-01-01 11:14 pm
參考: My Maths knowledge
2008-01-09 7:53 am
very Good
2008-01-01 11:36 pm
Let the two sequences be:
a, a+d, a+2d, ..., a+(n-1)d
and b, b+e, b+2e, ..., b+(n-1)e
Their sum:
[2a+(n-1)d]n/2
and [2b+(n-1)e]n/2
So, [2a+(n-1)d]/[2b+(n-1)e] = (2+3n)/(3+2n)
Since the above equation is true for all n, we can sub. n to be any number.
Sub. n = 1, 2a/2b = 1 , so a = b
Sub. n = 2, (2a+d)/(2b+e) = 8/7
14a + d = 16a + 8e
2a + 8e = d
(d/a) - 8(e/a) = 2
9(d/a) - 72(e/a) = 18 --- (1)
Sub. n = 3, (2a+2d)/(2b+2e) = 11/9
9a + 9d = 11a + 22e
2a + 22e = 9d
9(d/a) - 22(e/a) = 2 --- (2)
(2) - (1): 50 (e/a) = -16
e/a = -8/25, sub. to (1)
d/a = 2 + 8(-8/25) = - 14/25
The ratio of their seventh terms
= (a+6d)/(b+6e)
= (a+6d)/(a+6e)
= (1+6d/a)/(1+6e/a)
= [25+6x(-14)]/[25+6x(-8)]
= 59/23
Try your best and do your best!
a, a+d, a+2d, ..., a+(n-1)d
and b, b+e, b+2e, ..., b+(n-1)e
Their sum:
[2a+(n-1)d]n/2
and [2b+(n-1)e]n/2
So, [2a+(n-1)d]/[2b+(n-1)e] = (2+3n)/(3+2n)
Since the above equation is true for all n, we can sub. n to be any number.
Sub. n = 1, 2a/2b = 1 , so a = b
Sub. n = 2, (2a+d)/(2b+e) = 8/7
14a + d = 16a + 8e
2a + 8e = d
(d/a) - 8(e/a) = 2
9(d/a) - 72(e/a) = 18 --- (1)
Sub. n = 3, (2a+2d)/(2b+2e) = 11/9
9a + 9d = 11a + 22e
2a + 22e = 9d
9(d/a) - 22(e/a) = 2 --- (2)
(2) - (1): 50 (e/a) = -16
e/a = -8/25, sub. to (1)
d/a = 2 + 8(-8/25) = - 14/25
The ratio of their seventh terms
= (a+6d)/(b+6e)
= (a+6d)/(a+6e)
= (1+6d/a)/(1+6e/a)
= [25+6x(-14)]/[25+6x(-8)]
= 59/23
Try your best and do your best!
收錄日期: 2021-04-25 15:23:51
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