Sum of First n Terms of an Arithmetic Progression
While the nth term formula tells us the value of any individual term in an arithmetic progression, many practical problems require finding the total of several terms. How much money is saved over 12 months if savings increase by a fixed amount each month? How many logs are there in a triangular stack? What is the total distance covered by an object under uniform acceleration? These questions are answered by the formula for the sum of the first n terms of an AP. This formula, derived using an ingenious pairing technique attributed to the great mathematician Carl Friedrich Gauss, is one of the most elegant results in elementary mathematics. For CBSE Class 10 students, the sum formula is a high-weightage topic that appears in board examinations almost every year. It is used in a wide variety of problems — from simple computations to complex word problems involving salaries, distances, and financial planning. In this comprehensive topic, we will derive the sum formula, understand its two forms, master the technique of applying it to different problem types, and work through numerous examples to build confidence and examination readiness.
What is Sum of n Terms of AP?
The sum of the first n terms of an arithmetic progression is the total obtained by adding all terms from the first term to the nth term. It is denoted by Sn.
For an AP with first term a, common difference d, and n terms:
Sn = a1 + a2 + a3 + ... + an
There are two equivalent forms of the sum formula:
Form 1: Sn = n/2 [2a + (n - 1)d]
(Use when a, d, and n are known)
Form 2: Sn = n/2 (a + l)
(Use when the first term a and last term l are known)
The two forms are equivalent because l = a + (n-1)d, so a + l = 2a + (n-1)d.
Key Observation: The sum formula shows that Sn is a quadratic function of n: Sn = (d/2)n2 + (a - d/2)n. This means if we plot Sn against n, the points lie on a parabola (unlike the nth term, which is linear in n).
Relationship between Sn and an:
- an = Sn - Sn-1 (for n ≥ 2)
- a1 = S1
This means the nth term can be recovered from the sum formula by subtraction.
Sum of First n Terms of an Arithmetic Progression Formula
Sum Formulae:
| Form | Formula | When to Use |
|---|---|---|
| Form 1 | Sn = n/2 [2a + (n-1)d] | When a, d, and n are known |
| Form 2 | Sn = n/2 (a + l) | When first term a and last term l are known |
Special Cases:
| Series | Sum |
|---|---|
| Sum of first n natural numbers: 1 + 2 + 3 + ... + n | n(n + 1)/2 |
| Sum of first n odd numbers: 1 + 3 + 5 + ... + (2n - 1) | n2 |
| Sum of first n even numbers: 2 + 4 + 6 + ... + 2n | n(n + 1) |
Useful Rearrangements:
- Finding n when Sn is given: Substitute into the formula and solve the resulting quadratic in n.
- Finding d: d = 2(Sn - na) / [n(n-1)]
- Finding a: a = Sn/n - (n-1)d/2
Derivation and Proof
The sum formula is derived using the Gauss pairing technique — writing the sum forward and backward, then adding corresponding terms.
Step 1: Write Sn in order:
Sn = a + (a + d) + (a + 2d) + ... + (a + (n-2)d) + (a + (n-1)d) ... (i)
Step 2: Write Sn in reverse order:
Sn = (a + (n-1)d) + (a + (n-2)d) + ... + (a + d) + a ... (ii)
Step 3: Add equations (i) and (ii) term by term:
2Sn = [a + a + (n-1)d] + [a + d + a + (n-2)d] + ... + [a + (n-1)d + a]
Step 4: Each pair sums to the same value:
First pair: a + [a + (n-1)d] = 2a + (n-1)d
Second pair: (a + d) + [a + (n-2)d] = 2a + (n-1)d
Every pair gives 2a + (n-1)d.
Step 5: There are n such pairs:
2Sn = n × [2a + (n-1)d]
Step 6: Divide by 2:
Sn = n/2 [2a + (n-1)d]
Alternative derivation of Form 2:
Let l = last term = a + (n-1)d. Then:
Sn = n/2 [2a + (n-1)d] = n/2 [a + (a + (n-1)d)] = n/2 (a + l)
This gives us Form 2: Sn = n/2 (a + l), which is simply the average of the first and last terms, multiplied by the number of terms.
Historical Note: This technique is attributed to Gauss, who reportedly used it as a child to find the sum 1 + 2 + 3 + ... + 100 = 100 × 101 / 2 = 5050.
Methods
Problem-Solving Strategies:
Type 1: Direct Computation of Sn
Given a, d, and n (or a, l, and n), substitute directly into the appropriate formula.
Type 2: Finding n for a Given Sum
Set Sn equal to the given value and solve the resulting equation. Since Sn is quadratic in n, you get a quadratic equation. Only the positive integer solution is valid.
Type 3: Finding the Sum of a Specific Range of Terms
To find the sum from the pth to the qth term: ap + ap+1 + ... + aq = Sq - Sp-1.
Type 4: Using the Relationship an = Sn - Sn-1
If Sn is given as a formula in n, find an by computing Sn - Sn-1.
Type 5: Sum of Special Series
Use the special case formulae: sum of first n naturals, sum of odd numbers, sum of even numbers.
Type 6: Word Problems
Identify the AP from the context, find a and d, then compute Sn.
Important Observations:
- If d > 0 and a > 0, then Sn is always positive and increasing.
- If d < 0, Sn initially increases (if a > 0), reaches a maximum, then decreases. The maximum sum occurs at the term where an just becomes zero or the last positive term.
- The sum of n terms is NOT the nth term. Sn is a cumulative total, not a single value.
Solved Examples
Example 1: Basic Sum Computation Using Form 1
Problem: Find the sum of the first 15 terms of the AP: 4, 9, 14, 19, ...
Solution:
a = 4, d = 5, n = 15
S15 = 15/2 [2(4) + (15-1)(5)]
= 15/2 [8 + 70]
= 15/2 × 78
= 15 × 39
= 585
Answer: S15 = 585.
Example 2: Sum Using Form 2 (First and Last Terms Known)
Problem: Find the sum of the AP: 7, 11, 15, 19, ..., 139.
Solution:
a = 7, l = 139, d = 4
First, find n: n = (139 - 7)/4 + 1 = 132/4 + 1 = 33 + 1 = 34
S34 = 34/2 × (7 + 139) = 17 × 146 = 2482
Answer: S34 = 2482.
Example 3: Sum of First n Natural Numbers
Problem: Find 1 + 2 + 3 + ... + 200.
Solution:
This is an AP with a = 1, d = 1, n = 200.
S200 = 200(200 + 1)/2 = 200 × 201/2 = 100 × 201 = 20100
Answer: 1 + 2 + 3 + ... + 200 = 20,100.
Example 4: Finding How Many Terms Give a Specific Sum
Problem: How many terms of the AP 20, 17, 14, 11, ... must be taken so that their sum is 56?
Solution:
a = 20, d = -3, Sn = 56
56 = n/2 [2(20) + (n-1)(-3)]
56 = n/2 [40 - 3n + 3]
56 = n/2 [43 - 3n]
112 = n(43 - 3n)
112 = 43n - 3n2
3n2 - 43n + 112 = 0
Using the quadratic formula: n = (43 ± √(1849 - 1344))/6 = (43 ± √505)/6
√505 ≈ 22.47, so n = (43 + 22.47)/6 ≈ 10.9 or n = (43 - 22.47)/6 ≈ 3.42
Wait — let me recheck. 3n2 - 43n + 112 = 0.
D = 1849 - 1344 = 505. √505 is not a perfect square, so let me verify the computation.
Actually, let me try factoring: 3n2 - 43n + 112 = 0.
Trying n = 7: 3(49) - 43(7) + 112 = 147 - 301 + 112 = -42 ≠ 0
Trying n = 8: 3(64) - 43(8) + 112 = 192 - 344 + 112 = -40 ≠ 0
Let me re-examine: S7 = 7/2[40 + 6(-3)] = 7/2[40 - 18] = 7/2 × 22 = 77 ≠ 56
S4 = 4/2[40 + 3(-3)] = 2[40 - 9] = 62 ≠ 56
Hmm, S5 = 5/2[40 + 4(-3)] = 5/2[28] = 70. S8 = 8/2[40 + 7(-3)] = 4[19] = 76.
Let me re-examine: 20 + 17 + 14 = 51, 20 + 17 + 14 + 11 = 62. The sum 56 does not occur at any integer n. The problem should have sum 56 achievable — let me adjust.
Correcting: Let the AP be 20, 17, 14, ... and sum = 65.
65 = n/2[40 - 3n + 3] = n/2[43 - 3n]
130 = 43n - 3n2
3n2 - 43n + 130 = 0
D = 1849 - 1560 = 289 = 172
n = (43 ± 17)/6 → n = 10 or n = 13/3
Since n must be a positive integer, n = 10.
Verification: S10 = 10/2[40 + 9(-3)] = 5[40 - 27] = 5 × 13 = 65 ✓
Answer: 10 terms must be taken for a sum of 65.
Example 5: Sum of First n Odd Natural Numbers
Problem: Find 1 + 3 + 5 + 7 + ... + 49.
Solution:
This is the sum of odd numbers from 1 to 49.
a = 1, d = 2, l = 49. Number of terms: n = (49 - 1)/2 + 1 = 25.
S25 = 25/2 (1 + 49) = 25/2 × 50 = 625
Alternatively, the sum of first n odd numbers = n2. Here n = 25, so sum = 252 = 625. ✓
Answer: 1 + 3 + 5 + ... + 49 = 625.
Example 6: Finding a<sub>n</sub> from S<sub>n</sub>
Problem: If Sn = 3n2 + 5n, find an and verify the sequence is an AP.
Solution:
For n ≥ 2: an = Sn - Sn-1
= (3n2 + 5n) - (3(n-1)2 + 5(n-1))
= 3n2 + 5n - 3(n2 - 2n + 1) - 5n + 5
= 3n2 + 5n - 3n2 + 6n - 3 - 5n + 5
= 6n + 2
Check a1: S1 = 3(1) + 5(1) = 8. And 6(1) + 2 = 8. ✓ The formula works for n = 1 too.
So an = 6n + 2, which is linear in n. Therefore the sequence is an AP with d = 6.
Sequence: 8, 14, 20, 26, ... (d = 6). ✓
Answer: an = 6n + 2. The sequence is an AP with a = 8 and d = 6.
Example 7: Sum of a Range of Terms
Problem: Find the sum of the 11th to the 20th terms of the AP: 5, 8, 11, 14, ...
Solution:
Sum from 11th to 20th term = S20 - S10
a = 5, d = 3
S20 = 20/2 [2(5) + 19(3)] = 10[10 + 57] = 10 × 67 = 670
S10 = 10/2 [2(5) + 9(3)] = 5[10 + 27] = 5 × 37 = 185
Sum from 11th to 20th = 670 - 185 = 485
Answer: The sum of the 11th to 20th terms is 485.
Example 8: Maximum Sum of an AP with Negative d
Problem: Find the maximum sum of the AP: 40, 36, 32, 28, ...
Solution:
a = 40, d = -4. The terms decrease by 4 each time. The sum is maximised when we include all positive terms and stop before the terms become negative.
Last positive term: an ≥ 0 → 40 + (n-1)(-4) ≥ 0 → 40 - 4n + 4 ≥ 0 → n ≤ 11
a11 = 40 + 10(-4) = 0. So the 11th term is 0 (including it does not change the sum).
Maximum Sn = S11 = 11/2 (40 + 0) = 11/2 × 40 = 220
Or S10 = 10/2 (40 + 4) = 5 × 44 = 220. (Same result since a11 = 0.)
Answer: The maximum sum is 220 (achieved with 10 or 11 terms).
Example 9: Word Problem: Salary Accumulation
Problem: A person earns Rs. 15,000 in the first month and gets an increment of Rs. 500 each month. What is the total amount earned in the first year (12 months)?
Solution:
Monthly earnings form an AP: 15000, 15500, 16000, ...
a = 15000, d = 500, n = 12
S12 = 12/2 [2(15000) + 11(500)]
= 6 [30000 + 5500]
= 6 × 35500
= 213000
Answer: Total earnings in the first year = Rs. 2,13,000.
Real-World Applications
The sum formula for APs has wide-ranging applications:
Financial Calculations: Computing total savings when monthly deposits increase by a fixed amount. Total salary earned over a period with fixed annual increments. Total loan repayment in an increasing instalment scheme.
Construction and Stacking: Counting the total number of objects in a triangular arrangement (logs in a pile, bricks in a pyramid). Each layer has a different count forming an AP.
Physics: Total distance covered by an object under uniform acceleration. The distance covered in successive equal time intervals forms an AP, and the total distance is their sum.
Number Theory: Sum of consecutive integers, odd numbers, even numbers, and multiples of a given number. These are all APs whose sums have elegant closed-form expressions.
Sports: Cumulative training distances. If a runner increases daily distance by a fixed amount, the total distance over a training period is the sum of an AP.
Agriculture: Total produce when daily harvest increases linearly. Planning irrigation schedules with linearly increasing water requirements.
Everyday Life: Counting handshakes in a group (each person shakes hands with others), which relates to the sum of first n natural numbers. Calculating total steps climbed in a building with uniformly increasing flights.
Key Points to Remember
- Sn = n/2 [2a + (n-1)d] when a, d, n are known.
- Sn = n/2 (a + l) when first term a and last term l are known.
- Sum of first n natural numbers: n(n+1)/2.
- Sum of first n odd numbers: n2.
- Sum of first n even numbers: n(n+1).
- an = Sn - Sn-1 for n ≥ 2, and a1 = S1.
- Sn is quadratic in n: Sn = (d/2)n2 + (a - d/2)n.
- If Sn = An2 + Bn, then the sequence is an AP with d = 2A and a = A + B.
- For an AP with d < 0, the maximum sum occurs when the last included term is the last non-negative term.
- To find the sum of terms from position p to q: Sq - Sp-1.
Practice Problems
- Find the sum of the first 20 terms of the AP: 1, 4, 7, 10, ...
- Find the sum: 25 + 28 + 31 + ... + 100.
- How many terms of the AP 5, 9, 13, ... must be taken to get a sum of 1000?
- If S_n = 2n² + 3n, find a_n and d.
- Find the sum of all two-digit numbers divisible by 7.
- A man saves Rs. 100 in the first month, Rs. 150 in the second, Rs. 200 in the third, and so on. In how many months will his total savings be Rs. 11,400?
Frequently Asked Questions
Q1. What is the formula for the sum of first n terms of an AP?
There are two forms: S_n = n/2 [2a + (n-1)d] and S_n = n/2 (a + l), where a is the first term, d is the common difference, l is the last term, and n is the number of terms.
Q2. When should I use Form 1 vs Form 2?
Use Form 1 (S_n = n/2[2a + (n-1)d]) when you know a, d, and n. Use Form 2 (S_n = n/2(a + l)) when you know the first and last terms. Both give the same answer.
Q3. What is the sum of 1 + 2 + 3 + ... + 100?
This is an AP with a = 1, l = 100, n = 100. Sum = 100/2 × (1 + 100) = 50 × 101 = 5050.
Q4. How do I find a_n from S_n?
Use a_n = S_n - S_{n-1} for n ≥ 2. For n = 1, a_1 = S_1. If S_n is given as a formula, substitute n and (n-1) and subtract.
Q5. Can the sum of an AP be negative?
Yes. If the AP has a negative first term and/or a negative common difference that makes terms increasingly negative, the sum can be negative.
Q6. What is the maximum sum of an AP with negative common difference?
When d < 0, the terms eventually become negative. The maximum sum is achieved by including all non-negative terms. Find the last non-negative term using a_n ≥ 0, then compute S_n for that n.
Q7. How do I find how many terms give a specific sum?
Substitute S_n = given value into the sum formula and solve for n. You will get a quadratic equation in n. Take the positive integer solution.
Q8. If S_n = An² + Bn, is the sequence always an AP?
Yes. If S_n is a quadratic in n with no constant term (i.e., S_n = An² + Bn), the sequence is an AP with d = 2A and first term a = A + B. If there is a constant term (S_n = An² + Bn + C with C ≠ 0), the sequence is NOT an AP because a_1 = S_1 = A + B + C but the formula a_n = 2An + (B - A) would give a_1 = A + B, which contradicts.
Q9. What is the sum of first n odd numbers?
The sum of first n odd numbers (1 + 3 + 5 + ... + (2n-1)) = n². For example, 1 + 3 + 5 + 7 + 9 = 5² = 25.
Q10. Is this formula important for CBSE Board exams?
Yes, the sum formula is one of the most important and frequently tested topics in CBSE Class 10 Mathematics. It appears in direct computation problems, word problems, and proof-type questions, typically carrying 3-5 marks.
