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等差数列中a1=3,a10=78,则a5+a6=?

题目详情
等差数列中a1=3,a10=78,则a5+a6=?
▼优质解答
答案和解析
等差数列中a1=3,a10=78,
d = (78-3) ÷ (10-1)
= 75 ÷ 9
= 3分之 25
则a5+a6= 2 (a 5.5)
= 2[ a1 + d(5.5 -1 )]
= 2 [ 3 + 25/3 x 4.5 ]
= 2 [ 3 + 37.5]
= 2 x 40.5
= 81