等差数列［an］的前n项和为Sn,已知a10=30,a20=50 (1)求通项an (2)令Sn=242,求n.

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等差数列［an］的前n项和为Sn,已知a10=30,a20=50 (1)求通项an (2)令Sn=242,求n.
等差数列［an］的前n项和为Sn,已知a10=30,a20=50 (1)求通项an (2)令Sn=242,求n.

等差数列［an］的前n项和为Sn,已知a10=30,a20=50 (1)求通项an (2)令Sn=242,求n.
(1) a10 = a+9d = 30
a20 = a+19d = 50
d=2,a = 12
an = 12 + 2(n-1) = 2n +10
(2) sn = 2* (1+n)*n/2 + 10n = n^2 + 11n = 242
n = 11

d=(a20-a10)/10=2
a1=a10-9d=12
an=12+(n-1)*2=2n+10
Sn=(12+2n+10)n/2=n^2+11n=242
n^2+11n-242=0
(n-11)(n+22)=0
n＝11

设d为公差，a20=a10+10d，d=2，a1=a10-9d，a1=12，an=2n+10。
Sn=n(a1+an)/2=n（11+n），n=11

a20=a10+(20-10)*d将数字代入得到d=2 an=a10+(n-10)*2得到an=2n+10 sn=n*(a1+an)/2=n平方+11n 将sn=242代入，得到n=22

a10=a1+9d
a20=a1+19d 解得d=2 a1=12 an=12+2(n-1)
sn=n(a1+an)/2
=n( a1+a1+2(n-1) )
242=n(12+12+2n-1)
解得 n=11

(1) a10 = a+9d = 30
a20 = a+19d = 50
∴ d=2, a = 12
∴ an = 12 + 2(n-1) = 2n +10
(2) sn = na1+ (1+n)*n*d/2
∴n^2 + 11n = 242
（n-11）(n+22)=0
n=11

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