Arithmetic Progressions
The nth Term
General Form of an AP: Any AP can be written as: \[ a,\ a+d,\ a+2d,\ a+3d,\ a+4d,\ \dots \] where \( a \) is the first term and \( d \) is the common difference.
nth Term Formula: The term at position \( n \) (called \( a_n \)) is given by: \[ a_n = a + (n-1)d \] This formula allows us to find any term without writing all previous terms.
Finding Missing Terms: If some terms of an AP are missing, we can use the fact that the middle term of three consecutive terms is the arithmetic mean: For three terms \( x, y, z \) in AP: \( y = \frac{x+z}{2} \)
Checking if a number belongs to an AP: To check if a number \( N \) is a term of an AP with first term \( a \) and common difference \( d \):
- Set \( N = a + (n-1)d \)
- Solve for \( n \). If \( n \) is a positive integer, \( N \) is a term of the AP.
GENERAL FORM OF AP
Position: 1st 2nd 3rd 4th 5th ... nth
Term: a → a+d → a+2d → a+3d → a+4d ... a+(n-1)d
+d +d +d +d
FINDING THE nth TERM (Formula in Action)
Example: AP = 4, 9, 14, 19, 24, ...
a = 4, d = 5
a₁ = 4 + (1-1)×5 = 4
a₂ = 4 + (2-1)×5 = 9
a₃ = 4 + (3-1)×5 = 14
a₁₀ = 4 + (10-1)×5 = 4 + 45 = 49
CHECKING IF A NUMBER BELONGS TO AN AP
Is 78 a term of AP: 6, 13, 20, 27, ...?
Step 1: a = 6, d = 7
Step 2: 78 = 6 + (n-1)×7
Step 3: 78 - 6 = (n-1)×7 → 72 = (n-1)×7 → n-1 = 72/7 = 10.285...
Step 4: n is NOT an integer → 78 is NOT a term
ARITHMETIC MEAN (Middle Term)
Three consecutive terms: ____ , 12 , 20
Middle term = (first + third)/2
12 = (first + 20)/2 → 24 = first + 20 → first = 4
AP: 4, 12, 20- First term \( a = 7 \), common difference \( d = 13-7 = 6 \)
- \( a_n = a + (n-1)d \)
- \( a_{15} = 7 + (15-1)×6 = 7 + 14×6 = 7 + 84 = 91 \)
Answer: The 15th term is 91.
- \( a = 100 \), \( d = 97-100 = -3 \)
- \( a_n = 100 + (n-1)(-3) = 76 \)
- \( 100 - 3(n-1) = 76 \)
- \( -3(n-1) = 76 - 100 = -24 \)
- \( n-1 = \frac{-24}{-3} = 8 \)
- \( n = 9 \)
Answer: 76 is the 9th term.
- \( a_8 = a + 7d = 37 \) …(1)
- \( a_{12} = a + 11d = 57 \) …(2)
- Subtract (1) from (2): \( 4d = 20 \) → \( d = 5 \)
- From (1): \( a + 35 = 37 \) → \( a = 2 \)
- AP: 2, 7, 12, 17, 22, …
Answer: AP is 2, 7, 12, 17, 22, …
- General form: \( a, a+d, a+2d, a+3d, \dots \)
- nth term formula: \( a_n = a + (n-1)d \)
- To find a missing term, use arithmetic mean: middle term = (sum of neighbors)/2
- To check if a number belongs to an AP, solve \( N = a + (n-1)d \); if \( n \) is a positive integer, it belongs.
- You can form an AP from two given terms by finding \( a \) and \( d \).
Sum of n Terms
What is Summation? The sum of the first \( n \) terms of an AP is denoted by \( S_n \). It is the total when you add all terms from the first to the nth term.
Formula for Sum of First n Terms: There are two common formulas: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] \[ S_n = \frac{n}{2} (a + l) \] where \( l = a_n = a + (n-1)d \) is the last term.
Which formula to use?
- Use \( S_n = \frac{n}{2}[2a + (n-1)d] \) when you know \( a, d, n \).
- Use \( S_n = \frac{n}{2}(a + l) \) when you know \( a, l, n \).
Sum of Finite AP: For a finite AP with \( n \) terms, the sum is always a finite number.
Real-life applications:
- Total savings after \( n \) months with fixed monthly increase
- Total number of seats in an auditorium with increasing rows
- Distance traveled in \( n \) seconds with constant acceleration
- Total production over several years with fixed annual increase
SUM OF FIRST n TERMS (Visual Representation)
AP: 2, 5, 8, 11, 14 (a=2, d=3, n=5)
Write terms forward: 2 + 5 + 8 + 11 + 14 = 40
Write terms backward: 14 + 11 + 8 + 5 + 2 = 40
Add column by column: 16 + 16 + 16 + 16 + 16 = 80
Each pair sum = a + l = 2 + 14 = 16
Number of pairs = n = 5
Total of both rows = n × (a + l) = 5 × 16 = 80
So S_n = 80 ÷ 2 = 40 ✓
APPLICATION: SEATING ARRANGEMENT
Row 1: 10 seats
Row 2: 13 seats
Row 3: 16 seats
Total seats in 8 rows:
a=10, d=3, n=8
S₈ = 8/2[2×10 + (8-1)×3]
= 4[20 + 21] = 4×41 = 164 seats
SUM FORMULAE QUICK REFERENCE
┌────────────────────────────────────────┐
│ S_n = n/2 [2a + (n-1)d] │
│ S_n = n/2 (a + l) where l = last term │
└────────────────────────────────────────┘- \( a = 5, d = 6, n = 20 \)
- \( S_n = \frac{n}{2}[2a + (n-1)d] \)
- \( S_{20} = \frac{20}{2}[2×5 + (20-1)×6] \)
- \( = 10[10 + 19×6] = 10[10 + 114] = 10×124 = 1240 \)
Answer: The sum is 1240.
- \( a = 24, d = 21-24 = -3, S_n = 78 \)
- \( \frac{n}{2}[2×24 + (n-1)(-3)] = 78 \)
- \( \frac{n}{2}[48 - 3n + 3] = 78 \)
- \( \frac{n}{2}[51 - 3n] = 78 \)
- Multiply by 2: \( n(51 - 3n) = 156 \)
- \( 51n - 3n^2 = 156 \) → \( 3n^2 - 51n + 156 = 0 \)
- Divide by 3: \( n^2 - 17n + 52 = 0 \)
- \( (n-4)(n-13) = 0 \) → \( n = 4 \) or \( n = 13 \)
- Both are valid. For \( n=4 \): 24+21+18+15=78; For \( n=13 \): sum is also 78 (terms become negative later)
Answer: 4 terms or 13 terms.
- Penalties form AP: 200, 250, 300, …
- \( a = 200, d = 50, n = 30 \)
- \( S_{30} = \frac{30}{2}[2×200 + (30-1)×50] \)
- \( = 15[400 + 29×50] = 15[400 + 1450] \)
- \( = 15 × 1850 = 27750 \)
Answer: Total penalty = ₹27,750.
- Sum of first \( n \) terms: \( S_n = \frac{n}{2}[2a + (n-1)d] \) or \( S_n = \frac{n}{2}(a + l) \)
- The second formula is useful when the last term is given.
- Sum is finite for a finite number of terms.
- Real-life uses include total savings, seating capacity, penalties, and production totals.
- To find \( n \) given \( S_n \), form and solve a quadratic equation.
Applications of AP
Special Results (Sums of Special Sequences):
1. Sum of first n natural numbers: \( 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} \) This is an AP with \( a = 1, d = 1 \).
2. Sum of first n even numbers: \( 2 + 4 + 6 + \dots + 2n = n(n+1) \) This is an AP with \( a = 2, d = 2 \).
3. Sum of first n odd numbers: \( 1 + 3 + 5 + \dots + (2n-1) = n^2 \) This is an AP with \( a = 1, d = 2 \).
Word Problem Strategies:
- Read the problem carefully and identify what forms the AP.
- Find the first term \( a \) and common difference \( d \).
- Determine whether the problem asks for a particular term (\( a_n \)) or a sum (\( S_n \)).
- Translate words like "each year increase by fixed amount" → \( d \)
- "after n years" → look for \( a_n \) or \( S_n \)
- "total after n days" → usually \( S_n \)
Consecutive integer problems: Three consecutive terms in AP can be taken as: \( a-d, a, a+d \) Four consecutive terms: \( a-3d, a-d, a+d, a+3d \) This simplifies algebra.
SPECIAL SUM FORMULAE (Visual Proof) SUM OF FIRST n NATURAL NUMBERS 1 + 2 + 3 + ... + n = n(n+1)/2 Example: n=4 → 1+2+3+4 = 10 Formula: 4×5/2 = 10 ✓ SUM OF FIRST n EVEN NUMBERS 2 + 4 + 6 + ... + 2n = n(n+1) Example: n=4 → 2+4+6+8 = 20 Formula: 4×5 = 20 ✓ SUM OF FIRST n ODD NUMBERS 1 + 3 + 5 + ... + (2n-1) = n² Example: n=4 → 1+3+5+7 = 16 Formula: 4² = 16 ✓ STRATEGY FOR WORD PROBLEMS ┌─────────────────────────────────────┐ │ PROBLEM → Identify a and d │ │ ↓ │ │ Find term? Use a_n = a+(n-1)d │ │ Find sum? Use S_n formula │ │ Find n? Solve quadratic │ └─────────────────────────────────────┘ TAKING CONSECUTIVE TERMS IN AP 3 terms: a-d, a, a+d (sum = 3a) 4 terms: a-3d, a-d, a+d, a+3d (sum = 4a) 5 terms: a-2d, a-d, a, a+d, a+2d (sum = 5a)
- Odd numbers: 1, 3, 5, …, 49
- This is AP with \( a = 1, d = 2 \)
- Find n: \( a_n = 1 + (n-1)2 = 49 \) → \( 1 + 2n - 2 = 49 \) → \( 2n - 1 = 49 \) → \( 2n = 50 \) → \( n = 25 \)
- Sum = \( n^2 = 25^2 = 625 \) (using sum of first n odd numbers formula)
Answer: 625.
- Let numbers be \( a-d, a, a+d \)
- Sum: \( (a-d) + a + (a+d) = 3a = 24 \) → \( a = 8 \)
- Product: \( (a-d)×a×(a+d) = a(a^2 - d^2) = 8(64 - d^2) = 440 \)
- \( 64 - d^2 = 55 \) → \( d^2 = 9 \) → \( d = 3 \) or \( d = -3 \)
- Numbers: 5, 8, 11 or 11, 8, 5
Answer: 5, 8, 11 (or 11, 8, 5).
- Distance between top and bottom rungs = 2.5 m = 250 cm
- Gap between rungs = 25 cm
- Number of gaps = \( 250/25 = 10 \) gaps → number of rungs = \( 10 + 1 = 11 \)
- Lengths: 45, ?, ?, …, 25 (AP with 11 terms)
- \( a = 45, l = 25, n = 11 \)
- Total length = \( S_{11} = \frac{n}{2}(a + l) = \frac{11}{2}(45 + 25) = \frac{11}{2}×70 = 11×35 = 385 \) cm
Answer: 385 cm of wood.
- Sum of first n natural numbers: \( \frac{n(n+1)}{2} \)
- Sum of first n even numbers: \( n(n+1) \)
- Sum of first n odd numbers: \( n^2 \)
- For word problems, identify \( a, d, n \) from the situation.
- Use \( a-d, a, a+d \) for three consecutive terms in AP to simplify algebra.
- Check if the problem asks for a term (\( a_n \)) or a sum (\( S_n \)).