IMOClass 10 › Arithmetic Progressions

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
Example 1: Find the 15th term of the AP: 7, 13, 19, 25, …
  1. First term \( a = 7 \), common difference \( d = 13-7 = 6 \)
  2. \( a_n = a + (n-1)d \)
  3. \( a_{15} = 7 + (15-1)×6 = 7 + 14×6 = 7 + 84 = 91 \)

Answer: The 15th term is 91.

Example 2: Which term of the AP: 100, 97, 94, 91, … is 76?
  1. \( a = 100 \), \( d = 97-100 = -3 \)
  2. \( a_n = 100 + (n-1)(-3) = 76 \)
  3. \( 100 - 3(n-1) = 76 \)
  4. \( -3(n-1) = 76 - 100 = -24 \)
  5. \( n-1 = \frac{-24}{-3} = 8 \)
  6. \( n = 9 \)

Answer: 76 is the 9th term.

Example 3: The 8th term of an AP is 37 and the 12th term is 57. Find the AP.
  1. \( a_8 = a + 7d = 37 \) …(1)
  2. \( a_{12} = a + 11d = 57 \) …(2)
  3. Subtract (1) from (2): \( 4d = 20 \) → \( d = 5 \)
  4. From (1): \( a + 35 = 37 \) → \( a = 2 \)
  5. AP: 2, 7, 12, 17, 22, …

Answer: AP is 2, 7, 12, 17, 22, …

Quick recap
  • 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 \( d \).
✓ Quick check
The nth term of an AP with first term a and common difference d is:
The general term is aₙ = a + (n − 1)d.
If the sum of the first n terms of a sequence is Sₙ = n², then the nth term is:
aₙ = Sₙ − Sₙ₋₁ = n² − (n − 1)² = 2n − 1.

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  │
   └────────────────────────────────────────┘
Example 1: Find the sum of the first 20 terms of the AP: 5, 11, 17, 23, …
  1. \( a = 5, d = 6, n = 20 \)
  2. \( S_n = \frac{n}{2}[2a + (n-1)d] \)
  3. \( S_{20} = \frac{20}{2}[2×5 + (20-1)×6] \)
  4. \( = 10[10 + 19×6] = 10[10 + 114] = 10×124 = 1240 \)

Answer: The sum is 1240.

Example 2: How many terms of the AP: 24, 21, 18, … must be taken so that their sum is 78?
  1. \( a = 24, d = 21-24 = -3, S_n = 78 \)
  2. \( \frac{n}{2}[2×24 + (n-1)(-3)] = 78 \)
  3. \( \frac{n}{2}[48 - 3n + 3] = 78 \)
  4. \( \frac{n}{2}[51 - 3n] = 78 \)
  5. Multiply by 2: \( n(51 - 3n) = 156 \)
  6. \( 51n - 3n^2 = 156 \) → \( 3n^2 - 51n + 156 = 0 \)
  7. Divide by 3: \( n^2 - 17n + 52 = 0 \)
  8. \( (n-4)(n-13) = 0 \) → \( n = 4 \) or \( n = 13 \)
  9. 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.

Example 3: A contract on construction job specifies a penalty for delay beyond a certain date: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, and so on. Find the total penalty if the work is delayed by 30 days.
  1. Penalties form AP: 200, 250, 300, …
  2. \( a = 200, d = 50, n = 30 \)
  3. \( S_{30} = \frac{30}{2}[2×200 + (30-1)×50] \)
  4. \( = 15[400 + 29×50] = 15[400 + 1450] \)
  5. \( = 15 × 1850 = 27750 \)

Answer: Total penalty = ₹27,750.

Quick recap
  • 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.
✓ Quick check
The sum of the first n terms of an AP is:
Sₙ = n/2 [2a + (n − 1)d].
The sum of the first n odd natural numbers is:
1 + 3 + 5 + … + (2n − 1) = n².

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)
Example 1: Find the sum of all odd numbers between 0 and 50.
  1. Odd numbers: 1, 3, 5, …, 49
  2. This is AP with \( a = 1, d = 2 \)
  3. Find n: \( a_n = 1 + (n-1)2 = 49 \) → \( 1 + 2n - 2 = 49 \) → \( 2n - 1 = 49 \) → \( 2n = 50 \) → \( n = 25 \)
  4. Sum = \( n^2 = 25^2 = 625 \) (using sum of first n odd numbers formula)

Answer: 625.

Example 2: Find three numbers in AP whose sum is 24 and product is 440.
  1. Let numbers be \( a-d, a, a+d \)
  2. Sum: \( (a-d) + a + (a+d) = 3a = 24 \) → \( a = 8 \)
  3. Product: \( (a-d)×a×(a+d) = a(a^2 - d^2) = 8(64 - d^2) = 440 \)
  4. \( 64 - d^2 = 55 \) → \( d^2 = 9 \) → \( d = 3 \) or \( d = -3 \)
  5. Numbers: 5, 8, 11 or 11, 8, 5

Answer: 5, 8, 11 (or 11, 8, 5).

Example 3: A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2.5 m apart, find the total length of wood used for the rungs.
  1. Distance between top and bottom rungs = 2.5 m = 250 cm
  2. Gap between rungs = 25 cm
  3. Number of gaps = \( 250/25 = 10 \) gaps → number of rungs = \( 10 + 1 = 11 \)
  4. Lengths: 45, ?, ?, …, 25 (AP with 11 terms)
  5. \( a = 45, l = 25, n = 11 \)
  6. 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.

Quick recap
  • 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 (\( S_n \)).
✓ Quick check
A factory's output was 600 units in the first year and rose by 80 units each year. Its output in the 8th year was:
a₈ = 600 + 7 × 80 = ₹1160 units.
A child builds a triangle of coins: 1 in the top row, 2 in the next, and so on. To build 10 rows she needs:
1 + 2 + … + 10 = 10 × 11 ÷ 2 = 55 coins.
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