Sum of n Terms of an AP
1.0Master Sum of n Terms of an Arithmetic Progression in Minutes
2.0Learning Outcomes
After completing this topic, you will be able to:
- Understand the concept of the sum of an Arithmetic Progression.
- Apply the formula for the sum of the first n terms of an AP.
- Calculate the sum using the first term and common difference.
- Find the sum using the first and last terms.
- Solve application-based problems involving arithmetic progressions.
- Interpret real-life situations using AP concepts.
- Solve NCERT, competency-based, and CBSE Board examination questions confidently.
3.0Introduction to Sum of n Terms of an Arithmetic Progression
Have you ever calculated the total amount saved over several months when your savings increase by a fixed amount every month? Or found the total number of seats arranged in rows with a constant increase? These situations involve finding the sum of the terms of an Arithmetic Progression (AP).
Instead of adding each term individually, mathematics provides a simple formula to calculate the sum of the first n terms of an AP quickly and accurately. This concept has practical applications in finance, business, construction, sports scheduling, and data analysis. It is one of the most important and scoring topics in the CBSE Class 10 Mathematics syllabus.
4.0What Does the "Sum of First n Terms" Mean?
The arithmetic sequence or arithmetic progression is a series of numbers. You get more numbers by adding the same number every time you take one number from your arithmetic sequence or arithmetic progression.
The standard format of an AP looks like this:
a, a + d, a + 2d, a + 3d, ...
What we mean by finding the Sum of First n Terms, i.e., S or Sn, is the total obtained by adding together all the numbers from the first term to the nth term.
Sn = a + (a + d) + (a + 2d) + ... + [a + (n − 1)d]
5.0Derivation of the AP Sum Formula
Let S represent the total sum of the first n terms.
We can write the series as:
S = a + (a + d) + (a + 2d) + ... + [a + (n − 2)d] + [a + (n − 1)d] — (Equation 1)
Now write the same series in reverse order:
S = [a + (n − 1)d] + [a + (n − 2)d] + ... + (a + d) + a — (Equation 2)
Now add Equation 1 and Equation 2 term-wise.
First pair:
a + [a + (n − 1)d] = 2a + (n − 1)d
Second pair:
(a + d) + [a + (n − 2)d] = 2a + d + (n − 2)d = 2a + (n − 1)d
Last pair:
[a + (n − 1)d] + a = 2a + (n − 1)d
Notice that every pair adds up to the same value:
2a + (n − 1)d
Since there are n such pairs,
S + S = [2a + (n − 1)d] + [2a + (n − 1)d] + ... + [2a + (n − 1)d] (n times)
Therefore,
2S = n × [2a + (n − 1)d]
Dividing both sides by 2,
Sum of n Terms Formula (Standard Form)
S = n/2 × [2a + (n − 1)d]
The Alternate Form: Connecting to the Last Term (l)
We can simplify the formula further depending on the information given.
Split 2a as a + a: S = n/2 × [a + a + (n − 1)d]
Recall that the nth term of an AP is
an = a + (n − 1)d
If the AP has exactly n terms, then the nth term is also the last term, represented by l.
So,
l = a + (n − 1)d
Substituting this into the formula gives:
Last Term Sum Formula
S = n/2 × (a + l)
6.0Step-by-Step Solved Examples
Example 1: Finding the sum when a, d, and n are directly given
Question: Find the sum of the first 20 terms of the AP: 3, 8, 13, 18 …….
Solution:
Step 1: Extract the known values.
First term (a) = 3
Common difference (d) = 8 − 3 = 5
Number of terms (n) = 20
Step 2: Write the standard sum formula.
S = n/2 × [2a + (n − 1)d]
Step 3: Substitute the values.
S20 = 20/2 × [2 × 3 + (20 − 1) × 5]
= 10 × [6 + (19 × 5)]
= 10 × [6 + 95]
= 10 × 101
= 1010
Answer: The sum of the first 20 terms is 1,010.
Example 2: Finding the sum of a series when the last term is known
Question: Find the total sum of the arithmetic series: 5 + 12 + 19 + ... + 145
Solution:
Step 1: Identify the known values.
First term (a) = 5
Common difference (d) = 12 − 5 = 7
Last term (l) = 145
Step 2: Find the number of terms.
Using the nth term formula,
an = a + (n − 1)d
145 = 5 + (n − 1) × 7
145 − 5 = 7(n − 1)
140 = 7(n − 1)
n − 1 = 20
n = 21
Step 3: Use the last-term formula.
S = n/2 × (a + l)
S21 = 21/2 × (5 + 145)
= 21/2 × 150
= 21 × 75
= 1575
Answer: The sum of the series is 1,575.
Example 3: Finding missing variables given the sum
Question: The first term of an AP is 6 and the last term is 96. If the sum of all the terms is 816, find the total number of terms (n) and the common difference (d).
Solution:
Step 1: Identify the given values.a = 6, l = 96 and S = 816
Step 2: Use the last-term formula.
S = n/2 × (a + l)
816 = n/2 × (6 + 96)
816 = n/2 × 102
816 = 51n
n = 816 ÷ 51 = 16
Step 3: Find the common difference.
Using
l = a + (n − 1)d
96 = 6 + (16 − 1)d
96 − 6 = 15d
90 = 15d
d = 6
Answer:
Number of terms (n) = 16
Common difference (d) = 6
Example 4: Real-world word problem
Question: A professional decides to set up a monthly savings plan. They save ₹200 in the first month, ₹250 in the second month, ₹300 in the third month, and continue increasing their savings by ₹50 every month. What will be their total accumulated savings at the end of 2 years?
Solution:
Step 1: Convert the information into an AP.
Monthly savings: 200, 250, 300, ...
First term (a) = ₹200
Common difference (d) = ₹50
Time = 2 years = 24 months
Therefore, n = 24
Step 2: Apply the standard sum formula.
S = n/2 × [2a + (n − 1)d]
S24 = 24/2 × [2 × 200 + (24 − 1) × 50]
= 12 × [400 + (23 × 50)]
= 12 × [400 + 1150]
= 12 × 1550
= ₹18,600
Answer: The total accumulated savings at the end of 2 years will be ₹18,600..
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8.0Supporting Study Materials
This study material, including CBSE Notes and NCERT Solutions for the chapter Arithmetic Progressions on the topic Sum of n Terms of an Arithmetic Progression, is designed according to the latest CBSE Class 10 Mathematics syllabus and NCERT guidelines. It provides clear explanations of the formulas, derivations, solved examples, and application-based problems to help students understand how to calculate the sum of the first n terms of an AP efficiently. The study material also includes important board-level questions, competency-based exercises, and shortcut techniques to strengthen conceptual understanding and improve exam performance.
9.0Previous Year Question on Sum of an AP
Question: Find the sum of the AP: 3 + 7 + 11 + ... + 79.
Answer Given:
a = 3
d = 4
Last term (l) = 79
Find n: 79 = 3 + (n – 1) × 4
76 = 4(n – 1)
n = 20
Now, S = n/2 × (a + l)
= 20/2 × (3 + 79)
= 10 × 82
= 820
Read the passage and answer the questions.
Question: A theatre has 20 rows of seats. The first row contains 18 seats, and each subsequent row has 2 more seats than the previous one.
(i) Identify the type of sequence formed.
Answer: Arithmetic Progression
(ii) Find the number of seats in the last row.
l = 18 + 19 × 2
= 56 seats
(iii) Find the total number of seats in the theatre.
S = 20/2 × (18 + 56)
= 10 × 74
= 740 seats
(iv) Which formula is used to calculate the total number of seats?
Answer:
S = n/2 × (a + l)
10.030-Second Revision: Sum of an AP
- Arithmetic Progression has a constant common difference.
- Formula 1: Sₙ = n/2 × [2a + (n − 1)d]
- Formula 2:Sₙ = n/2 × (a + l)
- Use Formula 1 when a, d, and n are known.
- Use Formula 2 when a, l, and n are known.
- AP sums are widely used in finance, construction, scheduling, and everyday calculations.
11.0Recommended Next Topics