等差数列{an} ,a2+a7+a10=17,a4+a5+a6+...+a14=77,ak=13,则k=
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等差数列{an} ,a2+a7+a10=17,a4+a5+a6+...+a14=77,ak=13,则k=
设等差数列{an}的通项为:an = a1 +(n - 1)d,
其中a1为首项,d为公差,
由a2+a7+a10 = (a1 + d) + (a1 + 6d) + (a1 + 9d) = 3a1 + 16d,得:
3a1 + 16d = 17-----(1)
由a4+a5+a6+...+a14 = 1/2*(a4 + a14)*11 = 11*(a1 + 8d),得:
a1 + 8d = 7-----(2)
联立(1)(2)得:a1 = 3,d = 1/2.
所以an = 3 + 1/2(n - 1),
ak = 3 + 1/2(k - 1) = 13,
解得:k = 21.
其中a1为首项,d为公差,
由a2+a7+a10 = (a1 + d) + (a1 + 6d) + (a1 + 9d) = 3a1 + 16d,得:
3a1 + 16d = 17-----(1)
由a4+a5+a6+...+a14 = 1/2*(a4 + a14)*11 = 11*(a1 + 8d),得:
a1 + 8d = 7-----(2)
联立(1)(2)得:a1 = 3,d = 1/2.
所以an = 3 + 1/2(n - 1),
ak = 3 + 1/2(k - 1) = 13,
解得:k = 21.
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