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Find the first term of a sequence where the 31st, 32nd, and 33rd terms are 1.40, 1.55, and 1.70, respectively.

Answer :

jadeymae06, this is the solution:

This is an arithmetic sequence, where d (common difference) = 0.15

(1.70 - 1.55) or (1.55 - 1.40)

,

• a + 30d = 1.40

,

• a + 30(0.15) = 1.4

,

• a + 4.5 = 1.4

,

• a = 1.4 - 4.5

,

• a = -3.1

Jade, the first term is -3.1

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Rewritten by : Jeany