在等差数列an中已知a2+a6+a20+a24=48,求a13

2025-01-04 03:37:22
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回答1:

a1+d+a1+2d+a1+22d+a1+23d
=484a1+48d
=48a1+12d
=12a13=122)4a1+10d
=34,a1+d+a1+4d
=17(a1+d)(a1+4d)
=52
解得a1+d=13,
a1+4d=4或a1+d=4,a1+4d=13d=-3,或d=3.