nth Term of an Arithmetic Progression
The nth term formula is the most fundamental and widely used formula in the study of Arithmetic Progressions (AP). It provides a direct way to find any term of an AP without having to list all the preceding terms. If you know the first term and the common difference, you can jump directly to the 50th, 100th, or any term using this single formula. For CBSE Class 10 students, the nth term formula is the workhorse of the AP chapter — it is used in finding specific terms, determining whether a given number belongs to an AP, finding the number of terms in a finite AP, and solving word problems. Board examination questions on this formula range from straightforward computation to multi-step reasoning problems. The formula also connects beautifully to the concept of linear functions, as the nth term of an AP is a linear function of n. This topic will provide a thorough treatment of the nth term formula, including its derivation, interpretation, and a comprehensive set of examples covering every question type that students are likely to encounter.
What is nth Term of an AP?
The nth term (also called the general term) of an arithmetic progression gives the value of any term in the sequence based on its position number n. If the first term is a and the common difference is d, then the nth term is:
an = a + (n - 1)d
Here:
- an = the nth term (the term at position n)
- a = the first term (a1)
- n = the position number (n = 1, 2, 3, ...)
- d = the common difference
The formula tells us that to reach the nth term from the first term, we need to add the common difference (n - 1) times. The first term is at position 1, so we add d zero times. The second term adds d once. The third term adds d twice. And so on.
The nth term formula establishes that the terms of an AP form a linear function of n. Rewriting: an = dn + (a - d) = dn + (a - d). This is of the form an = pn + q, a linear expression in n. The slope is d (the common difference) and the y-intercept is (a - d). This linear nature means that if we plot the terms of an AP against their position numbers, the points lie on a straight line.
The last term of a finite AP is often denoted by l. If the AP has n terms, then l = a + (n - 1)d. This is particularly useful in the sum formula and in problems involving finite APs.
nth Term of an Arithmetic Progression Formula
nth Term Formula:
an = a + (n - 1)d
Useful Rearrangements:
| To Find | Rearranged Formula |
|---|---|
| nth term (an) | an = a + (n - 1)d |
| First term (a) | a = an - (n - 1)d |
| Common difference (d) | d = (an - a) / (n - 1) |
| Number of terms (n) | n = (an - a) / d + 1 |
nth Term from the End:
If an AP has last term l and common difference d, the nth term from the end is:
l - (n - 1)d
This is useful when you know the last term and need to count backwards.
Derivation and Proof
The nth term formula can be derived by observing the pattern of an AP:
Step 1: Write out the first few terms:
a1 = a (first term)
a2 = a + d (add d once)
a3 = a + 2d (add d twice)
a4 = a + 3d (add d three times)
a5 = a + 4d (add d four times)
Step 2: Observe the pattern:
For the 1st term: a + (1-1)d = a + 0d = a
For the 2nd term: a + (2-1)d = a + d
For the 3rd term: a + (3-1)d = a + 2d
For the 4th term: a + (4-1)d = a + 3d
Step 3: Generalise: The nth term adds d exactly (n-1) times to the first term:
an = a + (n - 1)d
Proof by Mathematical Induction (optional):
Base case: For n = 1, a1 = a + (1-1)d = a. True by definition.
Inductive step: Assume ak = a + (k-1)d for some k. Then ak+1 = ak + d = a + (k-1)d + d = a + kd = a + ((k+1)-1)d. This confirms the formula for k+1.
By mathematical induction, an = a + (n-1)d for all positive integers n.
Methods
Types of Problems Using the nth Term Formula:
Type 1: Finding a Specific Term
Given a and d, find an for a specific n. Directly substitute into an = a + (n-1)d.
Type 2: Checking Membership
Is a given number k a term of the AP? Set k = a + (n-1)d and solve for n. If n is a positive integer, k is a term of the AP. If n is not a positive integer, k is not in the AP.
Type 3: Finding the Number of Terms
Given the first term, last term, and common difference, find how many terms: n = (l - a)/d + 1.
Type 4: Finding a and d from Two Conditions
Given two pieces of information (like 'the 5th term is 23 and the 12th term is 58'), set up two equations and solve for a and d.
Type 5: nth Term from the End
The kth term from the end = l - (k-1)d, where l is the last term.
Type 6: Terms in AP Involving Variables
If terms are given in terms of k (like 2k+1, 3k, 4k-1), use the AP condition: 2(middle term) = first term + last term.
Type 7: Word Problems
Translate the verbal description into an AP context, identify a and d, then use the nth term formula to find the required value.
Solved Examples
Example 1: Finding a Specific Term
Problem: Find the 20th term of the AP: 3, 8, 13, 18, ...
Solution:
a = 3, d = 8 - 3 = 5, n = 20
a20 = a + (n - 1)d = 3 + (20 - 1)(5) = 3 + 19(5) = 3 + 95 = 98
Answer: The 20th term is 98.
Example 2: Checking if a Number Belongs to an AP
Problem: Is 301 a term of the AP: 5, 11, 17, 23, ...?
Solution:
a = 5, d = 11 - 5 = 6
Set 301 = 5 + (n-1)(6)
301 - 5 = (n-1)(6)
296 = 6(n-1)
n - 1 = 296/6 = 49.33...
Since n is not a positive integer, 301 is NOT a term of this AP.
Check: The 50th term = 5 + 49(6) = 299 and the 51st term = 305. Since 299 < 301 < 305, 301 falls between terms and is not in the AP.
Answer: No, 301 is not a term of this AP.
Example 3: Finding the Number of Terms
Problem: How many terms are in the AP: 7, 13, 19, ..., 205?
Solution:
a = 7, d = 6, l = 205
n = (l - a)/d + 1 = (205 - 7)/6 + 1 = 198/6 + 1 = 33 + 1 = 34
Answer: The AP has 34 terms.
Example 4: Finding a and d from Two Terms
Problem: The 7th term of an AP is 32 and the 13th term is 62. Find a and d, and write the AP.
Solution:
a7 = a + 6d = 32 ... (i)
a13 = a + 12d = 62 ... (ii)
Subtracting (i) from (ii): 6d = 30, so d = 5.
From (i): a + 30 = 32, so a = 2.
The AP is: 2, 7, 12, 17, 22, 27, 32, ...
Answer: a = 2, d = 5.
Example 5: Which Term is a Given Value?
Problem: Which term of the AP: 100, 97, 94, 91, ... is the first negative term?
Solution:
a = 100, d = -3
For the first negative term, an < 0:
100 + (n-1)(-3) < 0
100 - 3n + 3 < 0
103 < 3n
n > 34.33...
So the first negative term is the 35th term.
a35 = 100 + 34(-3) = 100 - 102 = -2
Verification: a34 = 100 + 33(-3) = 100 - 99 = 1 (positive). a35 = -2 (negative). ✓
Answer: The 35th term (-2) is the first negative term.
Example 6: nth Term from the End
Problem: Find the 5th term from the end of the AP: 3, 7, 11, ..., 103.
Solution:
l = 103, d = 4
5th term from the end = l - (5-1)d = 103 - 4(4) = 103 - 16 = 87
Verification: Total terms = (103-3)/4 + 1 = 100/4 + 1 = 26. The 5th from end = 26 - 5 + 1 = 22nd term = 3 + 21(4) = 87. ✓
Answer: The 5th term from the end is 87.
Example 7: Terms Equally Spaced from Ends
Problem: Show that in the AP 11, 18, 25, ..., 186, the sum of the 5th term from the start and the 5th term from the end equals the sum of the first and last terms.
Solution:
a = 11, d = 7, l = 186
5th term from start: a5 = 11 + 4(7) = 11 + 28 = 39
5th term from end: 186 - 4(7) = 186 - 28 = 158
Sum: 39 + 158 = 197
First + Last: 11 + 186 = 197
They are equal! ✓
This is a general property: in any AP, the sum of the kth term from the start and the kth term from the end equals (a + l) for all k.
Answer: 39 + 158 = 11 + 186 = 197. Property verified.
Example 8: Finding a Term When a<sub>n</sub> Is Given as a Formula
Problem: The nth term of a sequence is an = 3n - 7. Verify that it is an AP and find the 15th term.
Solution:
Check: a1 = 3(1) - 7 = -4, a2 = 3(2) - 7 = -1, a3 = 3(3) - 7 = 2, a4 = 3(4) - 7 = 5
Differences: -1-(-4) = 3, 2-(-1) = 3, 5-2 = 3. All equal. ✓ It is an AP with d = 3.
a15 = 3(15) - 7 = 45 - 7 = 38
Answer: Yes, it is an AP with d = 3. The 15th term is 38.
Example 9: Word Problem: Saving Money
Problem: Priya saves Rs. 200 in the first month and increases her savings by Rs. 50 each month. How much will she save in the 12th month?
Solution:
This forms an AP: 200, 250, 300, ... with a = 200, d = 50.
Savings in 12th month = a12 = 200 + (12-1)(50) = 200 + 550 = 750
Answer: Priya will save Rs. 750 in the 12th month.
Example 10: Finding the Term When Two APs Have Equal Terms
Problem: For what value of n will the nth term of the APs 63, 65, 67, ... and 3, 10, 17, ... be equal?
Solution:
AP 1: a = 63, d = 2. nth term = 63 + (n-1)(2) = 61 + 2n
AP 2: a = 3, d = 7. nth term = 3 + (n-1)(7) = -4 + 7n
Setting equal: 61 + 2n = -4 + 7n
65 = 5n
n = 13
Verification: AP 1: 61 + 26 = 87. AP 2: -4 + 91 = 87. Both give 87. ✓
Answer: The 13th terms of both APs are equal (both equal 87).
Real-World Applications
The nth term formula for APs has extensive real-world applications across many fields:
Financial Planning: Regular savings with fixed increments, salary structures with annual fixed raises, and instalment payment plans all follow AP patterns. The nth term formula predicts future values exactly. For example, if a person starts saving Rs. 500 in the first month and increases savings by Rs. 100 each month, the savings in the 24th month can be found directly: a24 = 500 + 23 × 100 = Rs. 2800.
Construction and Architecture: Staircases (each step has the same height), seating in auditoriums (rows with fixed increases in seats), and layered construction often follow AP patterns. If the first row has 20 seats and each subsequent row has 3 more, the nth term formula tells us exactly how many seats are in any row.
Physics: Objects under uniform acceleration cover distances in successive seconds that form an AP. Using the kinematic equation, the distance covered in the nth second is sn = u + (2n - 1)a/2, which is a linear function of n — an AP. This allows physicists to predict motion without tracking every second individually.
Sports and Fitness: Training schedules that increase intensity by fixed amounts use the AP formula to plan future workouts. A swimmer who increases laps by 2 each week from a starting count of 10 can use the formula to plan any future week: laps in week n = 10 + (n-1) × 2.
Computer Science: Loop counters, memory allocation in fixed-size blocks, and sequential file access patterns follow arithmetic sequences. Understanding AP patterns helps in algorithm analysis and predicting resource usage.
Calendar Calculations: Days of the week for the same date across successive months follow patterns that can be analysed using arithmetic sequences. Date calculations in software often use similar linear progression logic.
Agriculture: Planting patterns in rows with fixed spacing, and harvest planning based on linear growth models. If a field has rows with 12, 14, 16, ... plants, the nth term formula tells the farmer exactly how many plants go in any row.
Manufacturing: Production targets that increase by a fixed number each day, quality control sampling at regular intervals, and inventory management with fixed reorder quantities all use AP-based calculations.
Key Points to Remember
- The nth term of an AP is an = a + (n - 1)d.
- This formula gives any term directly from a, d, and n — no need to list all preceding terms.
- To check if k is a term: solve k = a + (n-1)d for n. If n is a positive integer, k is a term.
- Number of terms in a finite AP: n = (l - a)/d + 1.
- The nth term from the end: l - (n-1)d.
- If an = pn + q (linear in n), the sequence is an AP with d = p and first term a = p + q.
- A non-linear formula for an (like an = n2) does NOT give an AP.
- The formula can be rearranged to find a, d, or n when the other quantities are known.
- In any AP, the kth term from start + the kth term from end = first term + last term.
- The nth term formula is the basis for the sum formula Sn.
Practice Problems
- Find the 30th term of the AP: 10, 7, 4, 1, ...
- Is 184 a term of the AP: 3, 7, 11, 15, ...? If so, which term?
- How many terms are in the AP: 18, 15.5, 13, ..., -47?
- The 10th term of an AP is 52 and the 16th term is 82. Find the 32nd term.
- Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?
- Find the 8th term from the end of the AP: 2, 6, 10, ..., 86.
Frequently Asked Questions
Q1. What is the nth term of an AP?
The nth term (general term) of an AP with first term a and common difference d is aₙ = a + (n-1)d. It gives the value of any term at position n in the sequence.
Q2. How do I find which term of an AP equals a given number?
Set the given number equal to a + (n-1)d and solve for n. If n comes out as a positive integer, the number is the nth term. If n is not a positive integer, the number is not a term of the AP.
Q3. How do I find the number of terms in a finite AP?
Use n = (l - a)/d + 1, where l is the last term, a is the first term, and d is the common difference.
Q4. What is the nth term from the end of an AP?
The nth term from the end = l - (n-1)d, where l is the last term. Alternatively, if the AP has N terms, the nth term from the end is the (N - n + 1)th term from the start.
Q5. Can the nth term formula give a negative result?
Yes. If d is negative and n is large enough, or if the first term is negative, the nth term can be negative. For example, in the AP 10, 7, 4, 1, -2, ..., the 5th term is -2.
Q6. How is the nth term related to a linear function?
aₙ = a + (n-1)d = dn + (a-d). This is a linear function of n with slope d. The graph of term values vs position numbers is a straight line.
Q7. If aₙ = n² + 1, is the sequence an AP?
No. Check: a₁ = 2, a₂ = 5, a₃ = 10. Differences: 3 and 5 (not equal). Since aₙ is quadratic in n, not linear, the sequence is not an AP.
Q8. How do I find a and d if two terms are given?
Use the formula twice: a_m = a + (m-1)d and a_n = a + (n-1)d. Subtract to eliminate a and find d first, then substitute back to find a.
Q9. What is the difference between aₙ and Sₙ?
aₙ is the value of the individual nth term. Sₙ is the sum of the first n terms. They are related by aₙ = Sₙ - Sₙ₋₁ (for n ≥ 2).
Q10. Is this topic important for CBSE Board exams?
Yes, the nth term formula is one of the most frequently tested concepts in CBSE Class 10 Mathematics. Questions range from direct computation to multi-step word problems and typically carry 2-5 marks.
