Question

等差等比数列题

已知等差数列an 的公差和等比数列bn的公比都是d 且d不等于1 a1=b1 a4=b4 a10=b10 求 a1 和d b16是不是an 中的项？

Answer 1

你确认题目没有写错吧
等差数列 等比数列 前k项和
a(1) = a1 b(1) = b1
a(2) = a1 + d b(2) = b1*d
a(3) = a1 + 2d b(3) = b1*d²
...... ......
a(n) = a1 + (n-1)d b(n) = b1*d^(n-1)

根据题意：
a1=b1 ①
a(4) = b(4) ===> a1 + 3d = b1*d³ ②
a(10)=b(10) ===> a1 + 9d = b1*d^9 ③
联立解得：a1 = 2^(1/3) d = - 2^(1/3)
或者 a1 = 0 d = 0 -------- 不知道现在的教材是不是该舍去？你来定吧
b(16) = a1*d^15 = -2^(16/3)
假设有某一项 a(k+1) = b(16)
即 2^(1/3) (1 + k) = - 2^(16/3)
则 k +1 = - 2^5
k = - 33 不是自然数，所以b16 不是an 中的项

追问

大哥其实我想问的是
a1 + 3d = b1*d³
a1 + 9d = b1*d^9
这个方程组怎么解

追答

② a1* (d^3 - 1) = 3d
③ a1* (d^9 - 1) = 9d
相除：(d^9 - 1) = 3(d^3 - 1)
记 d^3 = x 则 d^9 = x^3

则 (x^3 - 1) = 3(x - 1)
(x - 1)(x² + x +1) = 3(x - 1)
(x² + x +1) = 3
(x² + x - 2) = 0
(x -1) (x + 2) = 0
下略

追问

小弟不才但我觉得

假设有某一项 a(k+1) = b(16)
即 2^(1/3) (1 + k) = - 2^(16/3)
则 k +1 = - 2^5

中的(1 + k) 应是（1-k）吧

追答

你是正确的，是我粗心了
重新补充
假设有某一项 a(k+1) = b(16)
a1 + kd = b1*d^15
2^(1/3) + k[-2^(1/3)] = 2^(1/3) * [-2^(1/3)]^15
即 2^(1/3) (1 - k) = - 2^(16/3)
则 1 - k = - 2^5 = - 32
k = 33 正是自然数，所以b16 的确是an 中的第34项