Question

在公差为d的等差数列an中，已知a1＝10，且4(a2＋1)∧2＝5a1a3。求d，an；若d＜0，求|a1|+…+|an|。

Answer 1

a(n) = 10 + (n-1)d.

4[a(2)+1]^2 = 5a(1)a(3) = 5*10*[10 + 2d] = 4[10 + d + 1]^2 ,
100(5+d) = 4(11+d)^2,
25(5+d) = 121 + 22d + d^2,
0 = d^2 - 3d - 4 = (d-4)(d+1),
d = 4或d = -1.

a(n) = 10 + 4(n-1) = 4n+6或a(n) = 10-(n-1) = 11-n.

d<0时，d=-1. a(n) = 11-n.
n<=11时，a(n) = 11-n >=0, |a(n)|=11-n = a(n),
n>=12时，a(n) = 11-n<0, |a(n)| = n - 11 = -a(n).

t(n) = |a(1)| + |a(2)| + ... + |a(n)|,
n<=11时，t(n) = a(1)+a(2)+...+a(n) = 11n - n(n+1)/2.
n>=12时，t(n) = a(1)+a(2)+...+a(11) - a(12) - a(13) - ... - a(n)
= 2t(11) - [a(1)+a(2)+...+a(n)] = 2[11*2-11*12/2] - [11n - n(n+1)/2]
= 11*4 - 11*12 - 11n + n(n+1)/2
= n(n+1)/2 - 11n - 88