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How do you find the 35th term of the sequence in which a1 equals 7 and the common difference is 7?
Wiki User
∙ 13y ago
a(1) = 7, a(2) = a(1) + 7, a(3) = a(2) + 7 which = a(1) + 7 + 7 which = 7 + 7 + 7.
There are as many multiples of 7 as there are terms, so a(35) = 7 x 35 = 245.
Wiki User
∙ 13y ago
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Find the 35th term of the sequence in which a1 equals 7 and the common difference is 7?
a(n) = 7n so a(35) = 245
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an = 4n - 14 So a35 = 35*4 - 14 = 140 - 14 = 126
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Find the 35th term of the sequence in which a1 equals 7 and the common difference is 7?
a(n) = 7n so a(35) = 245
What is the 35th term of the sequence in which a1 -10 and the common difference is 4?
an = 4n - 14 So a35 = 35*4 - 14 = 140 - 14 = 126
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