-8
-28
It is a + 8d where a is the first term and d is the common difference.
a + 99d where 'a' is the first term of the sequence and 'd' is the common difference.
From any term after the first, subtract the preceding term.
6
If the first term, t(1) = a and the common difference is r then t(n) = a + (n-1)*r where n = 1, 2, 3, ...
It is a + 8d where a is the first term and d is the common difference.
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
a + 99d where 'a' is the first term of the sequence and 'd' is the common difference.
100 - 13(4) = 48 or 100 + 13(4) = 152. (It was not stated whether the difference given is [term - preceding term] or [term - succeeding term]. * * * * * The common difference is defined as [term - preceding term] so the first answer is the correct one: 100 - 13*4 = 48
From any term after the first, subtract the preceding term.
If the first term is 12 and the seventh term is 36, then we have gone up 36-12 in the space of 6 term changes. This is 24 per 6 changes, which can be written as the division 24/6. This works out as 4. Thus the common difference in the sequence is 4.
The nth term is Un = a + (n-1)*d where a = U1 is the first term, and d is the common difference.
6
100 + 13(-4) = 100 - 56 = 48
6
29
-8