等差数列[an]的前n项和为Sn,已知a10=30,a20=50.(1)求通项公式an(2)若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)求通项公式an(2)若Sn=242,求n](/uploads/image/z/1819276-52-6.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%5Ban%5D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5a10%3D30%2Ca20%3D50.%281%29%E6%B1%82%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%282%29%E8%8B%A5Sn%3D242%2C%E6%B1%82n)
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等差数列[an]的前n项和为Sn,已知a10=30,a20=50.(1)求通项公式an(2)若Sn=242,求n
等差数列[an]的前n项和为Sn,已知a10=30,a20=50.(1)求通项公式an(2)若Sn=242,求n
等差数列[an]的前n项和为Sn,已知a10=30,a20=50.(1)求通项公式an(2)若Sn=242,求n
设公差为d
30=a1+(10-1)d
50=a1+(20-1)d
a1=12,d=2
(1)数列{an}的通项公式为
an=12+2(n-1)
即an=2n+10
(2)若Sn=242
n(12+2n+10)/2=242
n²+11n-242=0
n=11(n=-22不合题意舍去)
a20-a10=10d=20
d=2
a1=a10-9d=12
所以an=a1+2(n-1)=2n+10
Sn=(12+2n+10)*n/2=242
n²+11n-242=0
(n-11)(n+22)=0
所以n=11
10d=50-30=20
d=2
a1=30-9d=30-18=12
an=12+(n-1)*2=2n+10
sn=(12+2n+10)*n/2=242
n^2+11n-242=0
n=11
公差d等于a20减去a10除以(20-10),d=2, an=a10+2(n-10)=2n+10.
首项a1=2+10=12, Sn=(12+2n+10)n/2=242, 解出n=11
d=(a20 -a10)/(20-10) =2
an=2n+10
Sn=(2n+10+12)*n/2=242
n=11
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