Sn Notation:
= Sum of first n terms
= Replacement for Σ
= No need for Σ
EG: S10 = Sum of first 10 terms (10 Σ r=1)
= Replacement for Σ
= No need for Σ
EG: S10 = Sum of first 10 terms (10 Σ r=1)
General Arithmetic Sequence Rule:
a (a+d) (a+2d) (a+3d) ... U10 = (a+9d) Un = a+(n-1)d - In the Formula Book
+d +d +d
+d +d +d
- They all have a "common difference"
- a = the first term
- d = common difference
- n = number of terms
Proof of the Sn Formula:
Step 1: Write the first 3 terms and the last 3 terms of an arithmetic sequence in terms of (a+(n-1)d):
Sn = a+(a+d)+(a+2d)+...+(a+(n-3)d)+(a+(n-2)d)+(a+(n-1)d)
Step 2: Reverse the order of the Sum:
Sn = (a+(n-1)d)+(a+(n-2)d)+(a+(n-3)d)+...+(a+2d)+(a+d)+a
Step 3: Add the two sums from Step 1 and Step 2 together:
2Sn = (2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)
Step 4: Collect all the like terms:
2Sn = n(2a+(n-1)d)
Step 5: Get Sn on it's own:
Sn = n/2 (2a+(n-1)d) - In the Formula Book
Sn = a+(a+d)+(a+2d)+...+(a+(n-3)d)+(a+(n-2)d)+(a+(n-1)d)
Step 2: Reverse the order of the Sum:
Sn = (a+(n-1)d)+(a+(n-2)d)+(a+(n-3)d)+...+(a+2d)+(a+d)+a
Step 3: Add the two sums from Step 1 and Step 2 together:
2Sn = (2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)
Step 4: Collect all the like terms:
2Sn = n(2a+(n-1)d)
Step 5: Get Sn on it's own:
Sn = n/2 (2a+(n-1)d) - In the Formula Book
The Other Version:
As proved: Sn = n/2 (2a+(n-1)d)
Therefore: Sn = n/2 (a+(a+(n-1)d)) - with (a+(n-1)d) being the last term
Therefore: Sn = n/2 (a+L) - where L = Last Term - In the Formula Book
Therefore: Sn = n/2 (a+(a+(n-1)d)) - with (a+(n-1)d) being the last term
Therefore: Sn = n/2 (a+L) - where L = Last Term - In the Formula Book
Example:
Find the sum of the Arithmetic Sequence:
2 + 8 + 14 + ... + 284
a = 2 d = 6 n = ?
Find n using Un = a+(n-1)d
284 = 2+(n-1)6 (-2)
282 = (n-1)6 (/6)
47 = n-1 (+1)
48 = n
Therefore: a = 2 d = 6 n = 48
Substitute a, d and n into the Sn formula:
Sn = n/2 (2a+(n-1)d) or Sn = 1/2 n(a+L) - Use the 2nd version as the last term is known
S48 = 48/2 (2+284)
S48 = 24 (286)
S48 = 6864
2 + 8 + 14 + ... + 284
a = 2 d = 6 n = ?
Find n using Un = a+(n-1)d
284 = 2+(n-1)6 (-2)
282 = (n-1)6 (/6)
47 = n-1 (+1)
48 = n
Therefore: a = 2 d = 6 n = 48
Substitute a, d and n into the Sn formula:
Sn = n/2 (2a+(n-1)d) or Sn = 1/2 n(a+L) - Use the 2nd version as the last term is known
S48 = 48/2 (2+284)
S48 = 24 (286)
S48 = 6864