∵an为等差数列∴a2=a1+d,a3=a1+2d,a4=a1+3d又a2*a3=45∴(a1+d)(a1+2d)=45又a1+a4=14∴a1+(3/2)d=7a1+d=7-(1/2)d∴[7-(1/2)d]*[7+(1/2)d]=45d^2=16d=4,d=-4(舍去,∵d>0)a1=7-(3/2)*4 =1∴Sn=na1+[n(n-1)d/2] ...
已知等差数列{an}中,公差d>0,其前n项和为Sn,且满足:a2*a3=45,a1+a4=14 已知等差数列{an}中
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