数列{an},an=29,Sn=155,d=3,求a1

根据等差数列公式可得到：
an=a1+(n-1)d
29=a1+28d
a1=-55.

Sn=a1d+n(n-1)/2*d=3a1+3n(n-1)/2=155
Sn=(a1+an)n/2=(a1+29)n/2=155

a1可求

a(n) = a + 3(n-1),

S(n) = na + 3n(n-1)/2, n = 1,2,...

29 = a + 3(n-1) = a + 3n - 3,
a = 32 - 3n

155 = na + 3n(n-1)/2 = n(32-3n) + 3n(n-1)/2
= 32n - 3n^2 + 3n^2/2 - 3n/2
= 61n/2 - 3n^2/2,

0 = 3n^2 - 61n + 310 = (3n - 31)(n - 10),

n = 10.
a = 32 - 3n = 32 - 30 = 2.