Question

已知等差数列{An}中A10=30，A20=50。(1)求通项公式；(2)若Sn=242，求项数n 。

急用 急…

Answer 1

a20-a10 = 10d = 20，得d = 2
所以a1 = a10-9d = 12
通项an = a1+(n-1)d = 2n+10

Sn = a1+a2+...+an = n(a1+an)/2 = n(a1+a1+(n-1)d)/2 = n(12+12+2(n-1))/2 = n^2+11n = 242，即n^2+11n-242 = 0，(n-11)(n+22) = 0，n = 11

Answer 2

A20=A10+10d d=2 A1=A10-9d=12 An=A1+(n-1)d=12+2n-2=2n+10Sn=n(A1+An)/2=n(12+2n+10)/2=242 n�0�5+11n-242=0 (n+22)(n-11)=0 n>0 n=11

Answer 3

(1)a10=30,a20=50d=(a20-a10)/10=2a1=a10-9d=12则an=12+(n-1)2