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The nth term of a arithmetic sequence is given by: a{n} = a{1} + (n - 1)d

→ a{5} = a{1} + (5 - 1) × 3

→ a{5} = 4 + 4 × 3 = 16.

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Q: The first term is 4 and the common difference is 3. the fifth term of this sequence is?

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What is the 14th term in the arithmetic sequence in which the first is 100 and the common difference is -4? a14= a + 13d = 100 + 13(-4) = 48

16

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a + (n-1)d = last number where a is the first number d is the common difference.

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a + 99d where 'a' is the first term of the sequence and 'd' is the common difference.

35 minus 4 differences, ie 4 x 6 so first term is 11 and progression runs 11,17,23,29,35...

What is the 14th term in the arithmetic sequence in which the first is 100 and the common difference is -4? a14= a + 13d = 100 + 13(-4) = 48

It is a + 8d where a is the first term and d is the common difference.

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You subtract any two adjacent numbers in the sequence. For example, in the sequence (1, 4, 7, 10, ...), you can subtract 4 - 1, or 7 - 4, or 10 - 7; in any case you will get 3, which is the common difference.

It is a*r^4 where a is the first term and r is the common ratio (the ratio between a term and the one before it).

From any term after the first, subtract the preceding term.

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The common difference is 6; each number after the first equals the previous number minus 6.

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