Sequences Revision 1
Arithmetic Sequences
Enzo
Arithmetic Sequences An
arithmetic sequence, or arithmetic progression, is a sequence where each term is obtained by adding a constant to the previous term.
The
sequence goes up or down in equal steps.
The
constant is called the common difference.
E.g. 1+3+5+7+9 The common difference is 2. The formula for the nth term is: 2n -1
An arithmetic sequence is defined by:
U1 = 5 Un+1 = Un + 7 So, U2 = 5 + 7 = 12 And, U3 = 12 + 7 = 19 To find the 50th term quickly you would do: U50 = U1 + (49 x 7) = (5 + 343) = 348 is the 50th term.
An arithmetic sequence with the first term ‘a’ and common difference ‘d’ is defined recursively by:
U1 = a,
The nth term of the sequence I given by:
Un = a + (n – 1)d
E.g.(Ex 2c qn 2, page 258) a=2 and d=3 so the 10th term Is: 2 + (10 – 1) x 3 = 2 + 27 = 29
Un+1 = Un + d
The formula for the sum of a sequence is:
S = ½ n(a + l) = ½n(2a + (n–1)d)
E.g. (Ex 2c qn 4, page 258)
1+4+7+10… 36 terms. a=1, d=3 Last term = a +(n-1)d
= 1+ 35 x 3 = 106
Sum of 36 terms: Method 1 OR: ½36(2 x 1 +(36–1) x 3) = 18(2 + (35 x 3)) = 1926
Method 2 ½36(1+106) =1926
Enzo
Arithmetic Sequences An
arithmetic sequence, or arithmetic progression, is a sequence where each term is obtained by adding a constant to the previous term.
The
sequence goes up or down in equal steps.
The
constant is called the common difference.
E.g. 1+3+5+7+9 The common difference is 2. The formula for the nth term is: 2n -1
An arithmetic sequence is defined by:
U1 = 5 Un+1 = Un + 7 So, U2 = 5 + 7 = 12 And, U3 = 12 + 7 = 19 To find the 50th term quickly you would do: U50 = U1 + (49 x 7) = (5 + 343) = 348 is the 50th term.
An arithmetic sequence with the first term ‘a’ and common difference ‘d’ is defined recursively by:
U1 = a,
The nth term of the sequence I given by:
Un = a + (n – 1)d
E.g.(Ex 2c qn 2, page 258) a=2 and d=3 so the 10th term Is: 2 + (10 – 1) x 3 = 2 + 27 = 29
Un+1 = Un + d
The formula for the sum of a sequence is:
S = ½ n(a + l) = ½n(2a + (n–1)d)
E.g. (Ex 2c qn 4, page 258)
1+4+7+10… 36 terms. a=1, d=3 Last term = a +(n-1)d
= 1+ 35 x 3 = 106
Sum of 36 terms: Method 1 OR: ½36(2 x 1 +(36–1) x 3) = 18(2 + (35 x 3)) = 1926
Method 2 ½36(1+106) =1926
