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.
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
6
-8
a + (n-1)d = last number where a is the first number d is the common difference.
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
16
It is a + 8d where a is the first term and d is the common difference.
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.
6
From any term after the first, subtract the preceding term.
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).
The common difference is 6; each number after the first equals the previous number minus 6.
Oh, what a happy little question! Let's paint a picture with numbers. If 24 is the fifth term in a sequence of 10 numbers, we can see that each term is increasing by the same amount. By dividing 24 by 5, we find that each term is increasing by 4. So, the sequence would be 8, 12, 16, 20, 24, 28, 32, 36, 40, 44. Happy counting!
6