a3² = a1 * a7 = 1 * a7 = a7
即：a3² = a7.(1)
假設等差數(shù)列的公差為k,那么：
a3 = a1 + 2k = 1 + 2k .(2)
a7 = a1 + 6k = 1 + 6k .(3)
將(2)和(3)代入（1）,得到：
(1 + 2k)² = 1+6k
4k² + 4k + 1 = 1 + 6k
4k² - 2k = 0
4k(k-1/2) = 0
所以：k = 0或者k=1/2
因為公差不為0,所以k=1/2,
所以：an = a1 + (n - 1)k = 1 + (n - 1) / 2 = (n + 1) / 2
即：an = (n + 1) / 2第二問呢？② 設{an}的前n項和Sn，求數(shù)列{Sn/n}的前n項和TnSn = a1 + a2 + ... + an = a1 + (a1 + k) + (a1 + 2k) + ... (a1 + (n-1)k) = na1 + n(n-1)k/2 = n + n(n-1)/4 = n²/4 + 3n/4令Rn = Sn / n = n/4 + 3/4Tn = R1 + R2 + R3 + ... + Rn= (1/4 + 3/4) + (2/4 + 3/4) + ... + (n/4 + 3/4)= (1+2+ ... + n) / 4 + 3n/4= n(n+1)/8 + 6n/8 = (n² + 7n) / 8