设等差数列{an}的前n项和为Sn若a2,a4,a5也成等差数列，则s8等于

设等差数列{an}的公差为d,首项为a1,则:

an = a1 + (n-1)d      (1)

又因a2,a4,a5也成等差数列,设该数列公差为k,首项为b,则:

a2 = b

a4 = b + 2k

a5 = b + 3k        (2)

将(1)式代入(2)式得:

b = a1

2k = a4 - a2 = a1 + 3d - (a1 + d) = 2d

k = d

由(1)式得:a8 = a1 + 7d

又因a2,a4,a5也成等差数列,由(2)式得b=a1,k=d,代入Sn = (a1 + an)n/2得:

S8 = [a1 + (a1 + 7d)]8/2 = 8(a1 + 7d)/2 = 4a1 + 28d

由题意知s8的值,且a1的值也已知,可以解出d的值,进而求出整个等差数列{an}。

例如:

已知:a1 = 3,s8 = 100

求:{an}

解:

s8 = 4a1 + 28d = 100

28d = 100 - 4*3 = 88

d = 88/28 = 3

则等差数列{an}为:

a1 = 3

a2 = a1 + d = 3 + 3 = 6

a3 = a1 + 2d = 3 + 2*3 = 9

...

a8 = a1 + 7d = 3 + 7*3 = 24

即等差数列{an}为{3,6,9,...,24}