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In a set of four numbers, the first three are in an arithmetic sequence with a common difference of -12 and the last three are in a geometric sequence. Provided the first number equals the fourth, find the four numbers. 数学についての英文です。 よろしくお願いします。

noname#130345
noname#130345
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  • ky072
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回答No.1

4つの数があります。 先頭から3つの数は、公差-12の等差数列を成しています。 末尾から3つの数は、等比数列を成しています。 最初の数と最後の数が等しいという条件において、 この4つの数を求めなさい。 {16, 4, -8, 16} でしょうか。

noname#130345
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お礼

ありがとうございます。 おかげで助かりました。

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