Question

数列题 求解

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Answer 1

an= a1+(n-1)d
Sn = a1+a2+...+an
= [2a1+(n-1)d]n/2
√S3 = √[3(a1+d)]
√S2 =√(2a1+d)
√S1 =√a1
//
√S3+√S1 =2√S2
√[3(a1+d)] +√a1 =2√(2a1+d)
[3(a1+d)] +a1 +2√a1. √[3(a1+d)]=4(2a1+d)
2√a1. √[3(a1+d)]=4a1+d
12a1.(a1+d) = (4a1+d)^2
4(a1)^2 -4a1.d +d^2=0
(2a1-d)^2 =0
a1=d/2
//
√S2 -√S1=d
√(2a1+d)-√a1 =d
√(d+d)-√(d/2) =d
2√d-√d =√2d
√d =√2d
√d(√2√d -1) =0
d=0 or 1/4

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答案是1/2