等差数列通项:an=a1+(n-1)d
等差数列前n项和:Sn=na1+[n(n-1)/2]d
等比数列通项:an=a1q^(n-1)
等比数列前项和:Sn=a1(q^n-1)/(q-1)
易知 d = 3
由an = a1 + (n-1)d
得n = 112
所以 Sn = na1 + [n(n-1)/2]d
= 112 * 1 + [112*(112 - 1)/2]*3
= 18760
等差数列通项:an=a1+(n-1)d
等差数列前n项和:Sn=na1+[n(n-1)/2]d
等比数列通项:an=a1q^(n-1)
等比数列前项和:Sn=a1(q^n-1)/(q-1)
易知 d = 3
由an = a1 + (n-1)d
得n = 112
所以 Sn = na1 + [n(n-1)/2]d
= 112 * 1 + [112*(112 - 1)/2]*3
= 18760