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Anonymous users2024-01-24(1) It is known from: a5 + a8 = a1 + 4d + a1 + 7d = 2a1 + 11d = 15
a1=(15 - 11d)/2
then a4 = a1 + 3d = (15-11d) 2 + 3d = (15 - 5d) 2
a10=a1 + 9d=(15-11d)/2 + 9d=(15 + 7d)/2
a4a10=[(15-5d)/2]•[15+7d)/2](15-5d)(15+7d)=55•4
d-1)(7d+1)=0
d = 1 or d = -1 7
d>0
d=1, then a1=(15 - 11 1) 2=2
an=2 + n-1)•1=n+1
2)S5=[5(a1 + a5)]/2
Tn=20/2•3 + 20/3•4 + 20/(n+1)(n+2)
20[1/2•3 + 1/3•4 +.1/(n+1)(n+2)]
20[1/2 - 1/3 + 1/3 - 1/4 +.1/(n+1) -1/(n+2)]
20[1/2 - 1/(n+2)]
10n/(n+2)
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Anonymous users2024-01-23Don't think I'm in the toilet and you can look it up.
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