Selecting Terms in AP
When a problem involves finding three, four, or five terms in AP whose sum or product is given, choosing the right representation for the terms simplifies the calculation dramatically.
This technique is part of Chapter 5 — Arithmetic Progressions in the CBSE Class 10 syllabus. Instead of writing terms as a, a+d, a+2d, ..., a symmetric selection around the middle term eliminates unnecessary variables.
The idea is simple: if the sum of terms is given, choose terms so that the common difference cancels out when added. This gives the middle term immediately, reducing the problem to one unknown.
What is Selecting Terms in AP?
Definition: Selecting terms in AP means choosing a representation for the unknown AP terms that minimises the number of unknowns, making the equations easier to solve.
Standard selections:
| Number of Terms | Representation | Sum |
|---|---|---|
| 3 terms | a − d, a, a + d | 3a |
| 4 terms | a − 3d, a − d, a + d, a + 3d | 4a |
| 5 terms | a − 2d, a − d, a, a + d, a + 2d | 5a |
Key advantage: In all cases, the sum depends only on a (the middle term), since the d-terms cancel out symmetrically.
Important note for 4 terms: The common difference between consecutive terms in the 4-term selection is 2d, not d. So the actual common difference of the AP is 2d.
Selecting Terms in AP Formula
Three Terms in AP:
a − d, a, a + d (Sum = 3a)
Four Terms in AP:
a − 3d, a − d, a + d, a + 3d (Sum = 4a)
Five Terms in AP:
a − 2d, a − d, a, a + d, a + 2d (Sum = 5a)
Properties used:
- Sum of 3 terms: (a−d) + a + (a+d) = 3a → a = Sum/3
- Product of 3 terms: (a−d)(a)(a+d) = a(a² − d²)
- Sum of 4 terms: 4a → a = Sum/4
- Product of 4 terms: (a−3d)(a−d)(a+d)(a+3d) = (a² − 9d²)(a² − d²)
Derivation and Proof
Why does this symmetric selection work?
- In an AP with odd number of terms, there is a natural middle term.
- Terms equidistant from the middle have equal deviations: −d and +d, −2d and +2d, etc.
- When summed, these deviations cancel: (a−d) + (a+d) = 2a.
- Total sum = n × middle term = na.
- This gives a directly: a = Sum / n.
For 4 terms (even count):
- There is no single middle term. Instead, choose two middle terms as a − d and a + d.
- The outer two terms are a − 3d and a + 3d (gap of 2d between consecutive terms).
- Sum = (a−3d) + (a−d) + (a+d) + (a+3d) = 4a.
- The common difference of the AP is 2d (not d).
Why NOT use a, a+d, a+2d for 3 terms?
- If you use a, a+d, a+2d, the sum = 3a + 3d = 3(a+d). This involves both a and d, making the problem harder.
- The symmetric selection a−d, a, a+d gives sum = 3a, involving only a. Much simpler.
Types and Properties
Type 1: Sum of three terms given
- Use a−d, a, a+d. Find a from sum, then use the second condition to find d.
Type 2: Sum and product of three terms given
- Sum → a. Product → equation in d. Solve for d.
Type 3: Sum of four terms given
- Use a−3d, a−d, a+d, a+3d. Find a from sum. Use second condition for d.
Type 4: Sum of five terms given
- Use a−2d, a−d, a, a+d, a+2d. Find a from sum. Use second condition for d.
Type 5: Angles in AP
- Three angles of a triangle in AP: use (A−d), A, (A+d). Sum = 180°.
Methods
Steps to solve:
- Determine the number of terms (3, 4, or 5).
- Choose the appropriate symmetric representation from the table.
- Use the sum condition to find a directly: a = Sum / (number of terms).
- Use the second condition (product, sum of squares, specific ratios, etc.) to set up an equation in d.
- Solve for d.
- Substitute a and d to get the actual terms.
- If d can be positive or negative, you get two APs (one ascending, one descending) — both are valid unless the problem specifies order.
Common Mistakes:
- For 4 terms: forgetting that the common difference of the AP is 2d, not d.
- Assuming only positive d — both signs are usually valid.
- Using a, a+d, a+2d instead of the symmetric form when the sum is given (makes algebra harder).
Solved Examples
Example 1: Three Terms: Sum and Product Given
Problem: The sum of three numbers in AP is 27 and their product is 648. Find the numbers.
Solution:
Let the terms be a − d, a, a + d.
Using sum:
- 3a = 27 → a = 9
Using product:
- (9 − d)(9)(9 + d) = 648
- 9(81 − d²) = 648
- 81 − d² = 72
- d² = 9 → d = ±3
The numbers:
- If d = 3: 6, 9, 12
- If d = −3: 12, 9, 6
Answer: The three numbers are 6, 9, 12.
Example 2: Three Terms: Sum and Sum of Squares Given
Problem: Three numbers in AP have sum 15 and sum of their squares is 83. Find the numbers.
Solution:
Let the terms be a − d, a, a + d.
Using sum:
- 3a = 15 → a = 5
Using sum of squares:
- (5−d)² + 25 + (5+d)² = 83
- 25 − 10d + d² + 25 + 25 + 10d + d² = 83
- 75 + 2d² = 83
- 2d² = 8 → d² = 4 → d = ±2
The numbers:
- If d = 2: 3, 5, 7
- If d = −2: 7, 5, 3
Answer: The three numbers are 3, 5, 7.
Example 3: Angles of a Triangle in AP
Problem: The three angles of a triangle are in AP. The greatest angle is twice the smallest. Find all three angles.
Solution:
Let the angles be (a − d)°, a°, (a + d)°.
Using angle sum:
- 3a = 180 → a = 60°
Using given condition:
- Greatest angle = 2 × smallest
- (60 + d) = 2(60 − d)
- 60 + d = 120 − 2d
- 3d = 60 → d = 20°
The angles: 40°, 60°, 80°.
Verification: 40 + 60 + 80 = 180°. 80 = 2 × 40. Both conditions satisfied.
Answer: The angles are 40°, 60°, 80°.
Example 4: Four Terms in AP
Problem: The sum of four numbers in AP is 32 and the sum of their squares is 276. Find the numbers.
Solution:
Let the terms be a − 3d, a − d, a + d, a + 3d.
Using sum:
- 4a = 32 → a = 8
Using sum of squares:
- (8−3d)² + (8−d)² + (8+d)² + (8+3d)² = 276
- Expanding: (64 − 48d + 9d²) + (64 − 16d + d²) + (64 + 16d + d²) + (64 + 48d + 9d²) = 276
- 256 + 20d² = 276
- 20d² = 20 → d² = 1 → d = ±1
The numbers (d = 1): 5, 7, 9, 11.
Verification: Common difference = 2d = 2. Sum = 32. Sum of squares = 25 + 49 + 81 + 121 = 276.
Answer: The four numbers are 5, 7, 9, 11.
Example 5: Five Terms in AP
Problem: The sum of five numbers in AP is 25 and the sum of their squares is 165. Find the numbers.
Solution:
Let the terms be a − 2d, a − d, a, a + d, a + 2d.
Using sum:
- 5a = 25 → a = 5
Using sum of squares:
- (5−2d)² + (5−d)² + 25 + (5+d)² + (5+2d)² = 165
- Expanding: (25 − 20d + 4d²) + (25 − 10d + d²) + 25 + (25 + 10d + d²) + (25 + 20d + 4d²) = 165
- 125 + 10d² = 165
- 10d² = 40 → d² = 4 → d = ±2
The numbers (d = 2): 1, 3, 5, 7, 9.
Verification: Sum = 25. Sum of squares = 1 + 9 + 25 + 49 + 81 = 165.
Answer: The five numbers are 1, 3, 5, 7, 9.
Example 6: Three Terms: Ratio Condition
Problem: Three numbers in AP have sum 30. The ratio of the first to the third is 3:7. Find the numbers.
Solution:
Let the terms be a − d, a, a + d.
Using sum:
- 3a = 30 → a = 10
Using ratio condition:
- (10 − d) / (10 + d) = 3/7
- 7(10 − d) = 3(10 + d)
- 70 − 7d = 30 + 3d
- 10d = 40 → d = 4
The numbers: 6, 10, 14.
Verification: Sum = 30. Ratio 6:14 = 3:7.
Answer: The three numbers are 6, 10, 14.
Example 7: Three Terms: One Term Given
Problem: Three numbers in AP have sum 24. The middle number is 8. Find all three numbers.
Solution:
Let the terms be a − d, a, a + d.
Middle number = a = 8.
Using sum:
- 3(8) = 24. Consistent (no new info — sum confirms a = 8).
Without a second condition (product, ratio, etc.), there are infinitely many solutions for d. Any value of d gives a valid AP: (8−d), 8, (8+d).
For example: d = 3 gives 5, 8, 11. d = 5 gives 3, 8, 13.
Answer: The middle term is 8. Additional information is needed to determine d uniquely.
Example 8: Four Terms: Product of Extremes Given
Problem: The sum of four terms in AP is 20 and the product of the first and last terms is 7. Find the terms.
Solution:
Let the terms be a − 3d, a − d, a + d, a + 3d.
Using sum:
- 4a = 20 → a = 5
Using product of extremes:
- (5 − 3d)(5 + 3d) = 7
- 25 − 9d² = 7
- 9d² = 18 → d² = 2 → d = ±√2
The terms (d = √2): 5 − 3√2, 5 − √2, 5 + √2, 5 + 3√2.
Approximate values: 0.76, 3.59, 6.41, 9.24.
Verification: Sum ≈ 20. Product of extremes = (5 − 3√2)(5 + 3√2) = 25 − 18 = 7.
Answer: The terms are 5 − 3√2, 5 − √2, 5 + √2, 5 + 3√2.
Example 9: Consecutive Multiples as AP
Problem: Find three consecutive multiples of 7 whose sum is 63.
Solution:
Three consecutive multiples of 7 form an AP with common difference 7.
Let the terms be a − 7, a, a + 7.
Using sum:
- 3a = 63 → a = 21
The multiples: 14, 21, 28.
Verification: All multiples of 7. Sum = 63.
Answer: The three multiples are 14, 21, 28.
Example 10: Quadratic from AP Terms
Problem: Three numbers in AP have sum 21. The sum of the first and third numbers is 4 more than twice the second number minus 10. Find the numbers.
Solution:
Let the terms be a − d, a, a + d.
Using sum:
- 3a = 21 → a = 7
Using second condition:
- (7 − d) + (7 + d) = 2(7) − 10 + 4
- 14 = 14 − 10 + 4 = 8
- 14 = 8 — contradiction!
The condition is inconsistent as stated. Note that (a−d) + (a+d) = 2a always. The condition "sum of first and third = 2 × second" is always true for any AP.
Revised problem: Sum = 21 and product = 231. Then a = 7 and 7(49 − d²) = 231 → 49 − d² = 33 → d² = 16 → d = ±4.
Answer: The numbers are 3, 7, 11.
Real-World Applications
Simplifying AP Problems:
- Any time the sum of terms in AP is given, symmetric selection reduces the number of unknowns from 2 to 1.
Geometry:
- Interior angles of a polygon in AP — symmetric selection helps find each angle.
- Sides or diagonals of regular figures in AP.
Physics:
- Distances covered in successive seconds under uniform acceleration form an AP.
Competitive Exams:
- This technique is widely used in Olympiad, NTSE, and JEE problems involving AP terms.
Key Points to Remember
- For 3 terms in AP, use: a − d, a, a + d.
- For 4 terms in AP, use: a − 3d, a − d, a + d, a + 3d (common difference = 2d).
- For 5 terms in AP, use: a − 2d, a − d, a, a + d, a + 2d.
- In all cases, sum = n × a, so a = Sum / n.
- The second condition (product, sum of squares, ratio) gives an equation in d.
- Both d and −d give valid APs (one ascending, one descending).
- For 4 terms, the actual common difference of the AP is 2d, not d.
- This technique is much simpler than using a, a+d, a+2d, ... when the sum is given.
- Always verify the answer satisfies both given conditions.
- For angles in a triangle in AP: the middle angle is always 60° (since 180/3 = 60).
Practice Problems
- The sum of three numbers in AP is 33 and their product is 1287. Find the numbers.
- Three numbers in AP have sum 12 and sum of their cubes is 408. Find the numbers.
- The angles of a quadrilateral are in AP. The greatest angle is 120°. Find all four angles.
- Four numbers in AP have sum 28 and product of the first and last is 40. Find the numbers.
- Five numbers in AP have sum 50 and the first term is 2. Find all five numbers.
- The sum of three terms in AP is 36 and the ratio of first to third is 5:13. Find the terms.
- Three consecutive even numbers are in AP. Their sum is 42. Find them.
- The sum of four numbers in AP is 40 and the ratio of the product of the first and last to the product of the two middle terms is 2:3. Find the numbers.
Frequently Asked Questions
Q1. Why use symmetric selection instead of a, a+d, a+2d?
Symmetric selection (a−d, a, a+d) gives sum = 3a, directly yielding a = Sum/3. Using a, a+d, a+2d gives sum = 3a + 3d, which involves both unknowns and makes the algebra harder.
Q2. What is the common difference for the 4-term selection?
For a−3d, a−d, a+d, a+3d, the difference between consecutive terms is 2d. So the actual common difference of the AP is 2d, not d.
Q3. Can d be negative?
Yes. d < 0 gives a decreasing AP, and d > 0 gives an increasing AP. Both are valid unless the problem specifies ascending or descending order.
Q4. What if the sum is not given?
If the sum is not given, the symmetric selection offers no advantage. In that case, use the standard form a, a+d, a+2d, ... and work with the conditions provided.
Q5. How do you select terms for 6 numbers in AP?
Use a−5d, a−3d, a−d, a+d, a+3d, a+5d. Sum = 6a. Common difference = 2d.
Q6. If the three angles of a triangle are in AP, what is the middle angle?
The middle angle is always 60°. Sum = 3a = 180° → a = 60°.
Q7. Can this method be used for GP (geometric progression)?
A similar idea works for GP: use a/r, a, ar for three terms. The product = a³, so a = cube root of product. But this is a GP technique, not AP.
Q8. What if d² gives a negative value?
If d² < 0, there is no real solution. This means no AP with real terms satisfies the given conditions. The problem may have inconsistent data.
