In this chapter, you will learn
- —Understand and identify arithmetic progressions
- —Find the nth term of an arithmetic progression
- —Calculate the sum of n terms of an AP
- —Apply the arithmetic mean concept
- —Find specific terms in a progression
- —Solve word problems involving arithmetic progressions
- —Understand applications in financial planning and real scenarios
- —Analyze and verify arithmetic progressions
Introduction and Definition
Arithmetic Progression (AP): A sequence of numbers where the difference between consecutive terms is constant.
Notation:
• First term: a (or a₁)
• Common difference: d
• General term (nth term): aₙ = a + (n-1)d
• AP sequence: a, a+d, a+2d, a+3d, ...
Examples of Arithmetic Progressions:
- 2, 4, 6, 8, 10, ... (a=2, d=2)
- 5, 10, 15, 20, 25, ... (a=5, d=5)
- 100, 90, 80, 70, 60, ... (a=100, d=-10)
- 0, 1/2, 1, 3/2, 2, ... (a=0, d=1/2)
- 10, 10, 10, 10, ... (a=10, d=0) - constant sequence
Identifying an AP:
- Check if difference between consecutive terms is constant
- Calculate: d = a₂ - a₁ = a₃ - a₂ = a₄ - a₃ = ...
- If d is same throughout, it's an AP
Key Points:
- Common difference d can be positive (increasing), negative (decreasing), or zero (constant)
- An AP is completely defined by its first term a and common difference d
- There can be finite or infinite APs
Exam Tip
Always check if it's an AP before solving. Calculate d from first few terms. Remember: d = any term - previous term (should be same).
Common Mistake
Students don't verify that the difference is constant throughout. Always check at least 3 differences: a₂-a₁, a₃-a₂, a₄-a₃.
General Term (nth Term Formula)
The formula to find any term in an arithmetic progression without listing all previous terms.
General Term Formula:
aₙ = a + (n-1)d
where:
• aₙ = nth term
• a = first term
• n = position of term
• d = common difference
Derivation:
- First term: a₁ = a = a + 0·d
- Second term: a₂ = a + d = a + (2-1)d
- Third term: a₃ = a + 2d = a + (3-1)d
- nth term: aₙ = a + (n-1)d
Example 1: Find 10th term of 2, 5, 8, 11, ...
- a = 2, d = 5 - 2 = 3, n = 10
- a₁₀ = 2 + (10-1)×3 = 2 + 27 = 29
Example 2: Find 15th term of 100, 90, 80, 70, ...
- a = 100, d = 90 - 100 = -10, n = 15
- a₁₅ = 100 + (15-1)×(-10) = 100 - 140 = -40
Example 3: Which term of 5, 8, 11, 14, ... is 77?
- a = 5, d = 3, aₙ = 77
- 77 = 5 + (n-1)×3
- 72 = (n-1)×3
- n-1 = 24 → n = 25
- So 77 is the 25th term
Key Points:
- Formula works for any term without calculating all previous terms
- Used to find unknown first term, common difference, or term number
- Can be rearranged: a = aₙ - (n-1)d, d = (aₙ - a)/(n-1), n = (aₙ - a)/d + 1
Exam Tip
Memorize aₙ = a + (n-1)d well. Use it to find any term directly. For finding n, substitute known values and solve.
Common Mistake
Common error: using aₙ = a + nd instead of a + (n-1)d. Remember: there are (n-1) gaps between first and nth term, so we use (n-1)d.
Sum of n Terms
Formula to calculate the sum of first n terms of an arithmetic progression.
Sum Formula (Two Forms):
Sₙ = n/2 × [2a + (n-1)d]
or
Sₙ = n/2 × [a + l]
where l = aₙ (last term) = a + (n-1)d
Derivation:
Sₙ = a + (a+d) + (a+2d) + ... + [a+(n-1)d]
Writing in reverse:
Sₙ = [a+(n-1)d] + [a+(n-2)d] + ... + (a+d) + a
Adding both: 2Sₙ = n×[2a + (n-1)d]
Therefore: Sₙ = n/2 × [2a + (n-1)d]
Example 1: Find sum of first 10 terms of 2, 5, 8, 11, ...
- a = 2, d = 3, n = 10
- Sₙ = 10/2 × [2(2) + (10-1)×3]
- S₁₀ = 5 × [4 + 27] = 5 × 31 = 155
Example 2: Find sum using last term
For AP 1, 3, 5, 7, ..., find sum of first 8 terms
- a = 1, d = 2, n = 8
- Last term: l = 1 + (8-1)×2 = 1 + 14 = 15
- Sₙ = 8/2 × [1 + 15] = 4 × 16 = 64
Example 3: Sum of 1 + 2 + 3 + ... + 100
- a = 1, d = 1, n = 100, l = 100
- Sₙ = 100/2 × [1 + 100] = 50 × 101 = 5050
Key Points:
- Two forms of sum formula - use whichever is convenient
- Formula 1 uses first term and common difference
- Formula 2 uses first and last term (more intuitive)
- Sum of natural numbers: 1 + 2 + 3 + ... + n = n(n+1)/2
Exam Tip
Use Sₙ = n/2 × [a + l] when last term is known or easy to find. Use Sₙ = n/2 × [2a + (n-1)d] otherwise. Both give same answer.
Common Mistake
Students forget to divide by 2. It's n/2, not n. Also, be careful with (n-1)d term - don't confuse with n·d.
Arithmetic Mean
The arithmetic mean is a central value in an arithmetic progression with special properties.
Arithmetic Mean of Two Numbers:
If A is the arithmetic mean of a and b, then:
A = (a + b) / 2
Insertion of Arithmetic Means:
To insert n arithmetic means between two numbers a and b, we create an AP of (n+2) terms where:
- First term = a
- Last term = b
- Total terms = n + 2
- Common difference: d = (b - a) / (n + 1)
Example 1: Find AM of 10 and 20
- A = (10 + 20) / 2 = 15
- Check: 10, 15, 20 is an AP with d = 5 ✓
Example 2: Insert 5 arithmetic means between 2 and 20
- We need AP: 2, A₁, A₂, A₃, A₄, A₅, 20
- Total terms = 7, n + 2 = 7 (so n = 5)
- d = (20 - 2) / 6 = 18/6 = 3
- AP: 2, 5, 8, 11, 14, 17, 20
- The 5 means are: 5, 8, 11, 14, 17
Property:
- In an AP with odd number of terms, the middle term is the arithmetic mean of all terms
- The middle term also equals (first term + last term) / 2
Key Points:
- Arithmetic mean of two numbers equals their average
- If three numbers form an AP, middle one is AM of first and third
- Useful in dividing an interval into equal parts
Exam Tip
For inserting n means between a and b, use d = (b-a)/(n+1). The formula divides the gap into (n+1) equal parts.
Common Mistake
Students use d = (b-a)/n instead of (b-a)/(n+1). Remember: n means create (n+1) gaps, not n gaps.
Finding Specific Terms and Conditions
Techniques to find specific terms or verify properties of arithmetic progressions.
Problem Type 1: Find mth and nth terms, then find specific relationships
- Given: mth term = p, nth term = q
- Find: a and d
- Solution: aₘ = a + (m-1)d = p ... (1), aₙ = a + (n-1)d = q ... (2)
- Subtract: (n-m)d = q - p → d = (q-p)/(n-m)
- From (1): a = p - (m-1)d
Example: In an AP, 4th term = 13 and 7th term = 22. Find a, d, and 20th term.
- a₄ = 13, a₇ = 22
- d = (22 - 13) / (7 - 4) = 9/3 = 3
- a = 13 - (4-1)×3 = 13 - 9 = 4
- a₂₀ = 4 + (20-1)×3 = 4 + 57 = 61
Problem Type 2: Find if a number is in the AP
- Set aₙ = given number and solve for n
- If n is a positive integer, the number is in the AP
- If n is not an integer or is negative, the number is not in the AP
Example: Is 50 a term of 2, 5, 8, 11, ...?
- aₙ = 2 + (n-1)×3 = 50
- 3(n-1) = 48 → n-1 = 16 → n = 17
- Yes, 50 is the 17th term
Problem Type 3: Sum-related conditions
- If sum of m terms = p and sum of n terms = q
- Can find a and d by solving simultaneous equations
- Also can find: sum of (m+n) terms, sum of next m terms, etc.
Key Points:
- Any two conditions (like two different terms) can determine an AP completely
- Always verify answers by substitution
- For sum conditions, use both values of n to get two equations
Exam Tip
When given two conditions like 'mth term = p and nth term = q', always subtract equations to find d directly.
Common Mistake
Students get confused with two equations. Always subtract one from other to eliminate 'a' and find 'd' first, then use it to find 'a'.
Word Problems and Real-World Applications
Arithmetic progressions appear in many real-world scenarios involving uniform changes.
Common Real-World Problems:
- Savings/Investment: Fixed amount saved/invested each period
- Depreciation: Fixed amount of value lost each year
- Salary Increment: Fixed annual salary increase
- Seating Arrangements: Rows with increasing/decreasing seats
- Building Heights: Increasing/decreasing floor heights
- Temperature: Uniform temperature changes
Example 1: Salary Increment
Problem: A person's salary in first year is Rs. 50,000. Each year it increases by Rs. 5,000. What is the salary in the 10th year? Total salary earned in 10 years?
- a = 50,000, d = 5,000, n = 10
- 10th year salary: a₁₀ = 50,000 + 9×5,000 = 95,000
- Total in 10 years: S₁₀ = 10/2 × [50,000 + 95,000] = 5 × 145,000 = 725,000
Example 2: Depreciation
Problem: A machine worth Rs. 10,00,000 depreciates by Rs. 50,000 each year. After how many years will its value be Rs. 3,00,000?
- a = 10,00,000, d = -50,000, aₙ = 3,00,000
- 3,00,000 = 10,00,000 + (n-1)×(-50,000)
- -7,00,000 = -(n-1)×50,000
- n - 1 = 14 → n = 15 years
Steps to Solve Word Problems:
- Identify the first term 'a' (initial value)
- Identify common difference 'd' (uniform change)
- Identify what's asked: nth term, sum, or another relation
- Apply appropriate formula
- Verify answer makes sense in context
Key Points:
- Look for words indicating AP: "increases/decreases by fixed amount", "each year", "each month"
- First term is the starting value (not necessarily the 1st numbered item)
- Always check if answer is reasonable (e.g., negative salary doesn't make sense)
Exam Tip
In word problems, clearly state what 'a' and 'd' represent. Check that 'd' is indeed constant. Verify final answer in original problem context.
Common Mistake
Students misidentify first term or common difference. Always read problem carefully: first term is the initial value, not necessarily a₁ in a sequence.
Chapter Summary
Arithmetic Progressions are fundamental sequences with uniform changes. This chapter covers:
- Definition & Identification: Sequences with constant common difference
- General Term: aₙ = a + (n-1)d for finding any term
- Sum Formula: Sₙ = n/2 × [2a + (n-1)d] or Sₙ = n/2 × [a + l]
- Arithmetic Mean: Average of terms, insertion between numbers
- Finding Terms & Conditions: Using given information to find a, d, and specific terms
- Applications: Salary, depreciation, seating, investments, and real-world scenarios
Exam Focus: Finding nth term, sum of n terms, proving AP nature, inserting arithmetic means, word problems with practical applications.