问题(STPM)

zipp_882000

11)the sum of 3rd and 4th term of an AP is 18,the 7th term exceeds the 5th term by 14.Find the frist term and the common difference 。

12)The three consecutive terms of an AP have sum of 36 and a product of 1428 .Find the three terms

zipp_882000

11)the sum of 3rd and 4th term of an AP is 18,the 7th term exceeds the 5th term by 14.Find the frist term and the common difference 。

12)The three consecutive terms of an AP have sum of 36 and a product of 1428 .Find the three terms


13)Find the n th term of


8)Given that   is an arithmetic sequence with a  

find  

[ 本帖最后由 zipp_882000 于 28-10-2006 10:27 AM 编辑 ]

dunwan2tellu

11)the sum of 3rd and 4th term of an AP is 18,the 7th term exceeds the 5th term by 14.Find the frist term and the common difference 。

3rd term of AP : a+(3-1)d=a+2d
4th term : a+3d

Sum : (a+2d) + (a+3d) = 18 ....(i)

7th term - 5th term = 14 => (a+6d) - (a+4d) = 14 ....(ii)
.... Solve(i),(ii)...

12)The three consecutive terms of an AP have sum of 36 and a product of 1428 .Find the three terms

Let the three consecutive AP be , a-d , a , a+d

Sum : (a-d) + a + (a+d) = 36 ...(i)
Product : (a-d)a(a+d) = 1428 ....(ii)
.
.

13(i) 看 pattern 就应该知道 = log_e 3^n 了吧？不需要什么步骤。
(ii) 同理如上

8) 你先把 a1 和 a4 rasionalise , i.e

a1 = 5/(3+Sqrt[7]) = 5(3-Sqrt[7])/(9-7) = (15-5Sqrt[7])/2

a4 = 4/(2Sqrt[7]-5) = 4(2Sqrt[7]+5)/(28-25) = (8Sqrt[7]+20)/3

在 AP 里， a1 = a , a4 = a+3d ，a10 = a+9d

所以 a4-a1 = 3d = (8Sqrt[7]+20)/3 - (15-5Sqrt[7])/2 = A

那么 a10 = a + 3(3d) = a1 + 3A = .....

zipp_882000

谢谢，你救了我一命!!!