Arithmetic Progression (AP)
1 / 41. Identifying the Sequence
An Arithmetic Progression adds (or subtracts) the same number every time.
3, 7, 11, 15, ...
Find the parameters:
First Term ($a$):
Common Difference ($d$):
Try a Trickier One:
10, 8, 6, 4, ...
$d = $
Remember: $d = \text{Second Term} - \text{First Term}$Correct! $d$ can be negative.
2. Finding the $n$-th Term
Formula: $$ u_n = a + (n-1)d $$
Let's find the 20th term of the sequence: $3, 7, 11, \dots$
Here $a=3$, $d=4$, and $n=20$.
Step 1: Set up the formula
$$ u_{20} = 3 + (\dots) \times 4 $$
What is $(n-1)$?
Step 2: Calculate
$$ u_{20} = 3 + (19 \times 4) $$
$$ u_{20} = 3 + 76 $$
$$ u_{20} = \dots $$
Correct. The 20th term is 79.
3. The Sum Formula ($S_n$)
Formula: $$ S_n = \frac{n}{2} [ 2a + (n-1)d ] $$
Task: Find the sum of the first 10 terms of $5, 8, 11, \dots$
($a=5, d=3, n=10$)
Part A: Fill in the components
$\frac{n}{2} = \frac{10}{2}$
$2a = 2(5)$
$(n-1)d = 9 \times 3$
Part B: Final Calculation
$$ S_{10} = 5 \times [ 10 + 27 ] $$
$$ S_{10} = 5 \times 37 $$
Total Sum:
Perfect!
4. Exam Style: Unknown $a$ and $d$
Problem: The 3rd term is 10. The 7th term is 22. Find $a$ and $d$.
Step 1: Write equations
Term 3 ($n=3$): $a + 2d = 10$ --- (Eq 1)
Term 7 ($n=7$): $a + 6d = 22$ --- (Eq 2)
Term 3 ($n=3$): $a + 2d = 10$ --- (Eq 1)
Term 7 ($n=7$): $a + 6d = 22$ --- (Eq 2)
Step 2: Elimination (Eq 2 - Eq 1)
$$ (a - a) + (6d - 2d) = (22 - 10) $$
$$ 4d = 12 $$
Step 3: Solve for d
$$ d = \frac{12}{4} = \mathbf{3} $$
Step 4: Substitute back to find a
Using Eq 1: $a + 2(3) = 10$
$a + 6 = 10$
$\mathbf{a = 4}$
Sequence: 4, 7, 10, 13...
$a + 6 = 10$
$\mathbf{a = 4}$
Sequence: 4, 7, 10, 13...