设{An}为等差数列,公差d为正数,已知a2+a3+a4=15 又(a3-1)的平方=a2*a4
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设{An}为等差数列,公差d为正数,已知a2+a3+a4=15 又(a3-1)的平方=a2*a4
1.求a1与d
2.求数列{An}的前n项和Sn
1.求a1与d
2.求数列{An}的前n项和Sn
![设{An}为等差数列,公差d为正数,已知a2+a3+a4=15 又(a3-1)的平方=a2*a4](/uploads/image/z/3445574-14-4.jpg?t=%E8%AE%BE%EF%BD%9BAn%EF%BD%9D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%85%AC%E5%B7%AEd%E4%B8%BA%E6%AD%A3%E6%95%B0%2C%E5%B7%B2%E7%9F%A5a2%2Ba3%2Ba4%3D15+%E5%8F%88%EF%BC%88a3-1%EF%BC%89%E7%9A%84%E5%B9%B3%E6%96%B9%3Da2%2Aa4)
1.{An}为等差数列,所以2a3=a2+a4
a2+a3+a4=15=3a3 a3=5
(5-1)^2=16=(5-d)(5+d)
d=3 (-3舍掉)
所以a1=5-2d=-1 d=3
2.
an=a1+(n-1)*d=3n-4
Sn=3-4+6-4+9-4+...+3n-4
=3*(1+2+3+...n)-4n
=3*n*(n+1)/2-4n
=3/2*n^2-5/2*n
a2+a3+a4=15=3a3 a3=5
(5-1)^2=16=(5-d)(5+d)
d=3 (-3舍掉)
所以a1=5-2d=-1 d=3
2.
an=a1+(n-1)*d=3n-4
Sn=3-4+6-4+9-4+...+3n-4
=3*(1+2+3+...n)-4n
=3*n*(n+1)/2-4n
=3/2*n^2-5/2*n
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