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等差数列an中,若a4+a6+a8+a10+a12=120,则a10-1/2a12
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等差数列an中,若a4+a6+a8+a10+a12=120,则a10-1/2a12
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题目感觉少了些条件 等差数列 an 所以 a4+a6+a8+a10+a12 = 5 * a8 = 120
a8 = 24
假设差是d
则a10 = a1+9d 1/2a12= 1/2 *a1 + 11/2d
a10 - 1/2a12 = 1/2a1 + 7/2d = 1/2 * a8 = 12
a8 = 24
假设差是d
则a10 = a1+9d 1/2a12= 1/2 *a1 + 11/2d
a10 - 1/2a12 = 1/2a1 + 7/2d = 1/2 * a8 = 12
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