数列（an）各项的倒数组成一个等差数列,若a5＝√2－1,a7＝√2＋1,求a3

1/an = 1/a1 + (n-1)d
n=5
1/a5 = 1/a1 + 4d
1/(√2-1) = 1/a1 + 4d (1)
n=7
1/a7 = 1/a1 + 6d
1/(√2+1) = 1/a1 + 6d (2)
2(1) - (2)
a3=1/a1 + 2d
= 2/(√2-1) - 1/(√2+1)
= √2+3
再问： 谢谢啦