Arithmetic Mean of an AP
The arithmetic mean (AM) of an arithmetic progression is the average of two or more terms in an AP. In the simplest case, if three numbers a, b, c are in AP, then b is the arithmetic mean of a and c.
This concept is part of Class 10 CBSE Chapter 5 (Arithmetic Progressions). It provides a quick way to find a missing term when the terms on either side are known, and is used to verify whether three given numbers form an AP.
The arithmetic mean is closely related to the common difference. If a, b, c are in AP, then b − a = c − b, which gives b = (a + c)/2. This simple result has powerful applications in problem-solving.
What is Arithmetic Mean of an AP - Formula, Properties & Solved Examples?
Definition: The arithmetic mean of two numbers a and c is the number b such that a, b, c are in arithmetic progression.
b = (a + c) / 2
Equivalently: b is the AM of a and c if and only if b − a = c − b (equal common differences).
General Definition: If n arithmetic means m₁, m₂, ..., mₙ are inserted between two numbers a and c, then a, m₁, m₂, ..., mₙ, c form an AP with (n + 2) terms.
Key Properties:
- The AM of two numbers lies exactly halfway between them on the number line.
- The AM of any number of terms in AP equals the middle term (for odd number of terms) or the average of the two middle terms (for even number of terms).
- In an AP, every term is the AM of the terms equidistant from it on both sides.
Arithmetic Mean of an AP Formula
Arithmetic Mean of Two Numbers:
AM = (a + c) / 2
Inserting n Arithmetic Means Between a and c:
Common difference d = (c − a) / (n + 1)
The n means are: a + d, a + 2d, a + 3d, ..., a + nd
AM of n Terms of an AP:
AM = (first term + last term) / 2 = (a + l) / 2
Also:
- AM of all terms = Sum / Number of terms = Sₙ/n
- For an AP with n terms: AM = a + (n−1)d/2 = (2a + (n−1)d) / 2
Derivation and Proof
Derivation of AM = (a + c)/2:
- If a, b, c are in AP, then the common difference is the same between consecutive terms.
- b − a = c − b (definition of AP)
- 2b = a + c
- b = (a + c) / 2
Derivation of d when inserting n means:
- The sequence is: a, m₁, m₂, ..., mₙ, c
- This AP has (n + 2) terms total.
- Last term = a + (n + 1)d = c
- d = (c − a) / (n + 1)
Property: AM of equidistant terms
- In an AP: a₁, a₂, a₃, ..., aₙ
- For any term aₖ (where 1 < k < n): aₖ₋₁ = aₖ − d and aₖ₊₁ = aₖ + d
- (aₖ₋₁ + aₖ₊₁)/2 = (aₖ − d + aₖ + d)/2 = 2aₖ/2 = aₖ
- Therefore, every term is the AM of the terms equidistant from it.
Types and Properties
Applications of Arithmetic Mean in AP:
- Finding a missing middle term: If a, ?, c are in AP, the missing term = (a + c)/2.
- Checking if three numbers are in AP: a, b, c are in AP if and only if 2b = a + c.
- Inserting arithmetic means: Given two numbers, insert n AMs to form an AP.
- Finding average of an AP: AM of all terms = (first + last)/2.
- Selecting terms: When 3 terms of AP are needed, choose a − d, a, a + d (their AM is a).
Shortcut for selecting AP terms:
| Number of Terms | Choose as | AM |
|---|---|---|
| 3 | a − d, a, a + d | a |
| 4 | a − 3d, a − d, a + d, a + 3d | a |
| 5 | a − 2d, a − d, a, a + d, a + 2d | a |
Solved Examples
Example 1: Finding the Arithmetic Mean of Two Numbers
Problem: Find the arithmetic mean of 12 and 28.
Solution:
AM = (a + c) / 2:
- AM = (12 + 28) / 2 = 40/2 = 20
Verification: 12, 20, 28 → d = 8 throughout ✓
Answer: AM = 20
Example 2: Checking if Numbers Form an AP
Problem: Determine whether 7, 12, 17 are in AP.
Solution:
Check: Is the middle term the AM of the other two?
- AM of 7 and 17 = (7 + 17)/2 = 24/2 = 12
- Middle term = 12 ✓
Alternative check: 2 × 12 = 24 = 7 + 17 ✓
Answer: Yes, 7, 12, 17 are in AP (common difference = 5).
Example 3: Finding a Missing Term Using AM
Problem: If 3, x, 15 are in AP, find x.
Solution:
Since x is the AM of 3 and 15:
- x = (3 + 15)/2 = 18/2 = 9
Verification: 3, 9, 15 → d = 6 ✓
Answer: x = 9
Example 4: Inserting 3 Arithmetic Means Between Two Numbers
Problem: Insert 3 arithmetic means between 4 and 24.
Solution:
Given: a = 4, c = 24, n = 3 means
Total terms: n + 2 = 5
Common difference:
- d = (c − a)/(n + 1) = (24 − 4)/4 = 20/4 = 5
The 3 means:
- m₁ = 4 + 5 = 9
- m₂ = 4 + 10 = 14
- m₃ = 4 + 15 = 19
AP: 4, 9, 14, 19, 24 ✓
Answer: The three AMs are 9, 14, 19
Example 5: Sum of Three Terms in AP Given Their AM
Problem: Three numbers in AP have a sum of 27 and a product of 648. Find the numbers.
Solution:
Let the three terms be: a − d, a, a + d
Sum = 27:
- (a − d) + a + (a + d) = 27
- 3a = 27 → a = 9
Product = 648:
- (9 − d)(9)(9 + d) = 648
- 9(81 − d²) = 648
- 81 − d² = 72
- d² = 9 → d = ±3
The numbers are: 6, 9, 12 (or 12, 9, 6)
Answer: The three numbers are 6, 9, 12
Example 6: Finding AM of First n Terms of an AP
Problem: Find the arithmetic mean of the first 20 terms of the AP: 3, 7, 11, 15, ...
Solution:
Given: a = 3, d = 4, n = 20
Last term (20th term):
- a₂₀ = a + 19d = 3 + 76 = 79
AM = (first + last)/2:
- AM = (3 + 79)/2 = 82/2 = 41
Verification using sum: S₂₀ = (20/2)(3 + 79) = 10 × 82 = 820. AM = 820/20 = 41 ✓
Answer: AM = 41
Example 7: Inserting Single AM Between Two Numbers
Problem: Insert one arithmetic mean between −5 and 13.
Solution:
AM = (−5 + 13)/2 = 8/2 = 4
Verification: −5, 4, 13 → d = 9 ✓
Answer: The arithmetic mean is 4
Example 8: Finding Value of k Using AM Condition
Problem: If 2k + 1, 3k + 1, and 4k − 1 are in AP, find k.
Solution:
For AP: 2(middle term) = sum of first and third terms
- 2(3k + 1) = (2k + 1) + (4k − 1)
- 6k + 2 = 6k
- 2 = 0
This is a contradiction, which means no value of k makes these three terms an AP.
Let us recheck: For AP, middle = AM of other two:
- 3k + 1 = [(2k + 1) + (4k − 1)]/2 = (6k)/2 = 3k
- 3k + 1 = 3k → 1 = 0 (impossible)
Answer: No value of k satisfies the condition. These expressions cannot form an AP for any k.
Modified problem: If 2k + 1, 3k, and 4k − 1 are in AP, find k.
- 2(3k) = (2k + 1) + (4k − 1)
- 6k = 6k ✓ (true for all k)
These are always in AP for every value of k.
Example 9: Five Arithmetic Means Between 5 and 35
Problem: Insert 5 arithmetic means between 5 and 35.
Solution:
Given: a = 5, c = 35, n = 5
d = (35 − 5)/(5 + 1) = 30/6 = 5
The 5 means:
- m₁ = 5 + 5 = 10
- m₂ = 5 + 10 = 15
- m₃ = 5 + 15 = 20
- m₄ = 5 + 20 = 25
- m₅ = 5 + 25 = 30
Complete AP: 5, 10, 15, 20, 25, 30, 35 ✓
Answer: The five AMs are 10, 15, 20, 25, 30
Example 10: AM Property of Equidistant Terms
Problem: In the AP 2, 5, 8, 11, 14, 17, 20, verify that the 4th term is the AM of the 2nd and 6th terms.
Solution:
- 2nd term = 5
- 4th term = 11
- 6th term = 17
AM of 2nd and 6th terms:
- = (5 + 17)/2 = 22/2 = 11 = 4th term ✓
This confirms: Any term of an AP is the arithmetic mean of terms equidistant from it.
Real-World Applications
Applications of Arithmetic Mean:
- Statistics: The average (mean) of a data set is the arithmetic mean. This is the most commonly used measure of central tendency.
- Physics: Average velocity = (initial + final velocity)/2 when acceleration is constant — this is the AM of two velocities.
- Finance: Simple average returns, average monthly expenses, and installment calculations use arithmetic mean.
- Sports: A batsman's average, a team's average score — all use the AM concept.
- Interpolation: Inserting equally spaced values between two known values (used in tables, data analysis).
Key Points to Remember
- Arithmetic mean of a and c: AM = (a + c)/2.
- a, b, c are in AP if and only if 2b = a + c.
- To insert n AMs between a and c: d = (c − a)/(n + 1).
- AM of all terms of an AP = (first term + last term)/2.
- Every term of an AP is the AM of the terms equidistant from it.
- For 3 terms in AP, choose: a − d, a, a + d. Their sum = 3a and AM = a.
- For 4 terms in AP, choose: a − 3d, a − d, a + d, a + 3d (common difference = 2d).
- The AM always lies between the two given numbers.
- If AM of a and b equals AM of c and d, then a + b = c + d.
Practice Problems
- Find the arithmetic mean of 18 and 42.
- If 5, x, 17 are in AP, find x.
- Insert 4 arithmetic means between 2 and 32.
- Three numbers in AP have a sum of 33. The product of the first and third is 80. Find the numbers.
- Find the AM of the first 50 terms of the AP: 1, 3, 5, 7, ...
- If a, b, c are in AP and a + b + c = 18, a² + b² + c² = 126, find a, b, c.
- Verify that in the AP 3, 8, 13, 18, 23, 28, the 3rd term is the AM of the 1st and 5th terms.
Frequently Asked Questions
Q1. What is the arithmetic mean of two numbers?
The arithmetic mean of two numbers a and c is (a + c)/2. It is the number that, when placed between a and c, forms an arithmetic progression.
Q2. How do you check if three numbers are in AP?
Three numbers a, b, c are in AP if 2b = a + c, i.e., the middle term is the arithmetic mean of the other two. Equivalently, b − a = c − b (equal common differences).
Q3. How do you insert n arithmetic means between two numbers?
Find d = (c − a)/(n + 1), where a is the first number and c is the last. The n means are a + d, a + 2d, ..., a + nd.
Q4. What is the AM of all terms of an AP?
AM = (first term + last term)/2 = (a + l)/2. This equals Sₙ/n where Sₙ is the sum of n terms.
Q5. Why choose a − d, a, a + d for three terms in AP?
This symmetric form has two advantages: (1) the sum simplifies to 3a, directly giving the AM, and (2) the product involves (a² − d²), simplifying calculations significantly.
Q6. Is the arithmetic mean the same as average?
Yes. In everyday usage, 'average' usually means the arithmetic mean = sum of values / number of values. In statistics, other types of means exist (geometric, harmonic), but AM is the most common.
Q7. Can the AM be outside the range of the two numbers?
No. The AM of a and c always lies between a and c (inclusive). AM = (a + c)/2 is exactly the midpoint on the number line.
Q8. What is the difference between arithmetic mean and median?
In an AP, the arithmetic mean of all terms equals the median (middle term). In general data, the mean (sum/count) can differ from the median (middle value when arranged in order).
