设等差数列an的首项为1,公差1/2,且数列am,15,Sn与lgam,lg9,lgSn均为等差关系

an=1+(1/2)(n-1)=(n+1)/2
am=(m+1)/2
Sn=(a1+an)n/2=[1+(n+1)/2]n/2=n(n+3)/4
am,15,Sn成等差,则
am+Sn=2×15=30
lgam,lg9,lgSn成等差,则
am×Sn=9²=81
am,Sn是方程x²-30x+81=0的两根.
(x-3)(x-27)=0
x=3或x=27
am=3 Sn=27或am=27 Sn=3
am=3 Sn=27时,
(m+1)/2=3 m=5
n(n+3)/4=27
n²+3n-108=0
(n+12)(n-9)=0
n=-12(舍去)或n=9
am=27 Sn=3时,
(m+1)/2=27 m=53
n(n+3)/4=3
n²+3n=12
(n+ 3/2)²=39/4
n=(-3+√39)/2或n=(-3-√39)/2,不是正整数,舍去.
综上,得m=5 n=9