What is 5.4 Sum of First n Terms of an AP in NCERT Class 10 Mathematics Chapter 5?

5.4 Sum of First n Terms of an AP is the topic of Part 2 of Chapter 5 (Ch 5 — Arithmetic Progressions) in NCERT Class 10 Mathematics. This lesson builds intuition through examples, visuals, and graded practice.

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5.4 Sum of First n Terms of an AP

Q: What is Part 2 u2014 Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool in NCERT Class 10 Mathematics?

Part 2 u2014 Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool is a key concept covered in NCERT Class 10 Mathematics, Chapter 5: Arithmetic Progressions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

Q: How do I solve problems on Part 2 u2014 Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool step by step?

To solve problems on Part 2 u2014 Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty u2014 start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

Q: Is Part 2 u2014 Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool important for the Class 10 board exam?

Part 2 u2014 Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

Q: What mistakes should students avoid in Part 2 u2014 Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool?

Common mistakes in Part 2 u2014 Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Q: Where can I find more NCERT practice questions on Part 2 u2014 Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool?

End-of-chapter NCERT exercises for Part 2 u2014 Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 5, and solve at least one previous-year board paper to consolidate your understanding.

🎓 Class 10 Mathematics CBSE Theory Ch 5 — Arithmetic Progressions ⏱ ~15 min

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5.4 Sum of First n Terms of an AP

A famous story tells of a school teacher who, to keep 10-year-old Carl Friedrich Gauss busy, asked the class to add the numbers 1 through 100. Within moments Gauss answered 5050. How? He noticed that pairing the numbers from the ends inward always gave 101:

Fig 5.2: Gauss's pairing trick — the original idea behind the AP sum formula

Derivation of the Sum Formula

Derivation

Let \(S = a + (a+d) + (a+2d) + \cdots + (a+(n-1)d)\).
Write \(S\) in reverse: \(S = (a+(n-1)d) + (a+(n-2)d) + \cdots + a\).
Add the two row-wise. Each pair sums to \(\,2a + (n-1)d\,\), and there are \(n\) pairs.
Hence \(2S = n\bigl[2a+(n-1)d\bigr]\), giving
\[S_n = \tfrac{n}{2}\bigl[\,2a+(n-1)d\,\bigr].\]

Key Formulas

\[\boxed{\,S_n = \tfrac{n}{2}[2a+(n-1)d]\,}\qquad\boxed{\,S_n = \tfrac{n}{2}(a+l)\,}\] where \(l = a_n = a+(n-1)d\) is the last (nth) term.
Also, \(a_n = S_n - S_{n-1}\).

Sum of the first \(n\) natural numbers: \(1+2+\cdots+n = \tfrac{n(n+1)}{2}\) (put \(a=1,\ d=1\)).

Example 11 — Sum of 22 terms

Find the sum of the first 22 terms of the AP 8, 3, −2, −7, ...

\(a=8,\ d=-5,\ n=22\). \(S_{22} = \tfrac{22}{2}[2\cdot 8 + 21\cdot(-5)] = 11[16-105] = 11(-89) = -979\).

Example 12 — Finding a specific term sum

If the first term of an AP is 5, the last term is 45, and the sum is 400, find the number of terms.

\(S_n = \tfrac{n}{2}(a+l)\Rightarrow 400 = \tfrac{n}{2}(5+45) = 25n\Rightarrow n = 16\).
Also \(d = \frac{l-a}{n-1} = \frac{40}{15} = \tfrac{8}{3}\).

Example 13 — How many terms for a given sum?

How many terms of the AP 24, 21, 18, ... must be taken to give a sum of 78?

\(a=24,\ d=-3\). \(78 = \tfrac{n}{2}[48 + (n-1)(-3)] = \tfrac{n}{2}(51 - 3n)\).
So \(156 = 51n - 3n^2\Rightarrow 3n^2 - 51n + 156 = 0\Rightarrow n^2 - 17n + 52 = 0\).
\((n-4)(n-13)=0\Rightarrow n = 4\) or \(13\). Both are valid (because terms 5 through 13 happen to sum to 0).

Example 14 — Sum of first 1000 positive integers

\(S_{1000} = \tfrac{1000\cdot 1001}{2} = 500500\).

Example 15 — Sum of first \(n\) odd numbers

AP 1, 3, 5, 7, ..., \(a=1,\ d=2\). \(S_n = \tfrac{n}{2}[2+(n-1)\cdot 2] = \tfrac{n}{2}\cdot 2n = n^2\). Neat result: the sum of the first \(n\) odd numbers is \(n^2\).

Example 16 — Managerial annuity

A manager saves ₹5000 in the first year of service and increases his annual savings by ₹200 every year. In how many years will his total savings be ₹66,000?

\(a=5000,\ d=200,\ S_n = 66000\). \(66000 = \tfrac{n}{2}[10000+(n-1)\cdot 200] = n(5000 + 100(n-1)) = n(4900+100n)\).
\(100n^2 + 4900n - 66000 = 0 \Rightarrow n^2 + 49n - 660 = 0\). \((n-11)(n+60)=0\Rightarrow n = 11\) (reject −60). 11 years.

Example 17 — Finding \(a_n\) from \(S_n\)

The sum of the first \(n\) terms of an AP is given by \(S_n = 4n - n^2\). Find \(a\), \(a_2\), and \(a_{10}\).

\(a = S_1 = 4-1 = 3\). \(S_2 = 8-4 = 4\Rightarrow a_2 = S_2 - S_1 = 1\). So \(d = -2\). \(a_{10} = 3 + 9(-2) = -15\).

Example 18 — Bricks in a pile

A pile of logs has 20 logs in the bottom row, 19 in the next, 18 in the next, and so on. How many rows are there if 200 logs are used? Is \(S_n = 200\) possible with the same pattern?

\(a=20,\ d=-1\). \(200 = \tfrac{n}{2}[40-(n-1)] = \tfrac{n}{2}(41-n)\Rightarrow n^2 - 41n + 400 = 0\Rightarrow (n-16)(n-25)=0\).
\(n=16\) gives last row \(a_{16} = 5\) logs — valid. \(n=25\) would give \(a_{25} = -4\) — impossible. So 16 rows.

Example 19 — Prize money

In a school, a cash prize of ₹700 is to be given for seven categories with each successive prize being ₹20 less than the previous. Find the value of each of the seven prizes.

\(S_7 = 700\), \(d=-20,\ n=7\). \(700 = \tfrac{7}{2}[2a + 6(-20)] = \tfrac{7}{2}(2a-120)\Rightarrow 2a-120 = 200\Rightarrow a = 160\).
Prizes: ₹160, 140, 120, 100, 80, 60, 40.

In-text: Why did we reject \(n=-60\) in Example 16 and \(n=25\) in Example 18? Because \(n\) must be a positive integer and every physical term must make sense (positive log count in Ex. 18).

Activity: The Paired-Sum Proof

L4 Analyse

Materials: 20 small coins or counters, table, partner.

Predict: What is \(1+2+3+\cdots+20\)?

Place 1 coin in column 1, 2 in column 2, ..., 20 in column 20. This is a staircase totaling \(S\) coins.
Stack an identical staircase upside-down on top — now every column has exactly 21 coins and there are 20 columns.
Total coins = \(20 \times 21 = 420\). But this is twice your original, so \(S = 210\).
Generalise: for \(1+2+\cdots+n\), the paired total is \(n(n+1)\); dividing by 2 gives \(\tfrac{n(n+1)}{2}\).

This is exactly the reverse-and-add trick used to derive \(S_n = \tfrac{n}{2}(a+l)\). The paired sum \(a+l\) appears \(n\) times, and halving gives the total.

Exercise 5.3 (Selected)

Q1. Find the sum of the following APs: (i) 2, 7, 12, ..., to 10 terms. (ii) −37, −33, −29, ..., to 12 terms. (iii) 0.6, 1.7, 2.8, ..., to 100 terms. (iv) \(\tfrac{1}{15}, \tfrac{1}{12}, \tfrac{1}{10}, \ldots\) to 11 terms.

(i) \(a=2,d=5,n=10\): \(S=\tfrac{10}{2}[4+45]=245\).
(ii) \(a=-37,d=4,n=12\): \(S=6[-74+44]=6(-30)=-180\). Wait: \(S=\tfrac{12}{2}[2(-37)+11(4)]=6(-74+44)=-180\).
(iii) \(a=0.6,d=1.1,n=100\): \(S=50[1.2+99(1.1)]=50(1.2+108.9)=50(110.1)=5505\).
(iv) \(d=\tfrac{1}{12}-\tfrac{1}{15}=\tfrac{5-4}{60}=\tfrac{1}{60}\). \(S_{11}=\tfrac{11}{2}\bigl[\tfrac{2}{15}+10\cdot\tfrac{1}{60}\bigr]=\tfrac{11}{2}\bigl[\tfrac{2}{15}+\tfrac{1}{6}\bigr]=\tfrac{11}{2}\cdot\tfrac{4+5}{30}=\tfrac{11}{2}\cdot\tfrac{9}{30}=\tfrac{33}{20}\).

Q2. How many terms of the AP 9, 17, 25, ... must be taken to give a sum of 636?

\(a=9,d=8\). \(636=\tfrac{n}{2}[18+8(n-1)]=n(5+4n)\Rightarrow 4n^2+5n-636=0\). \(n=\tfrac{-5\pm\sqrt{25+10176}}{8}=\tfrac{-5\pm 101}{8}\). \(n=12\).

Q3. The first term of an AP is 5, the last term is 45, and the sum is 400. Find the number of terms and the common difference.

\(S_n=\tfrac{n}{2}(a+l)\Rightarrow 400=\tfrac{n}{2}(50)=25n\Rightarrow n=16\). \(d=\tfrac{45-5}{15}=\tfrac{40}{15}=\tfrac{8}{3}\).

Q4. The sum of the first 15 multiples of 8.

AP 8, 16, 24, ..., 120. \(S=\tfrac{15}{2}(8+120)=\tfrac{15}{2}\cdot 128=960\).

Q5. Find the sum of the odd numbers between 0 and 50.

AP 1, 3, 5, ..., 49 has 25 terms. Using \(S_n=n^2\): \(S_{25}=625\).

Competency-Based Questions

Scenario: Rina plans to save for a cycle. She deposits ₹100 in January, ₹150 in February, ₹200 in March, and continues the pattern by increasing each month's deposit by ₹50 over the previous month.

Q1. How much will she deposit in December (month 12)?

L3 Apply

(a) ₹600
(b) ₹650
(c) ₹700
(d) ₹750

(b) ₹650. \(a=100,d=50\); \(a_{12}=100+11(50)=650\).

Q2. Analyse her total savings by the end of the year.

L4 Analyse

\(S_{12}=\tfrac{12}{2}(100+650)=6(750)=4500\). ₹4,500. Since the terms grow linearly, the sum grows quadratically with \(n\).

Q3. A cycle costs ₹7200. Evaluate whether she can afford it with 15 months of saving, and if not, what \(d\) would allow her to reach ₹7200 in exactly 12 months keeping \(a=100\)?

L5 Evaluate

15 months: \(S_{15}=\tfrac{15}{2}[200+14(50)]=\tfrac{15}{2}(900)=6750\) — short by ₹450. For 12 months to hit ₹7200: \(7200=\tfrac{12}{2}[200+11d]=6(200+11d)\Rightarrow 200+11d=1200\Rightarrow d\approx 90.9\). So she must step up by about ₹91 each month.

Q4. Design a 10-month savings plan with a fixed first deposit of ₹200 and a total of exactly ₹5000. State \(d\) and the monthly deposits.

L6 Create

\(5000=\tfrac{10}{2}[400+9d]=5(400+9d)\Rightarrow 400+9d=1000\Rightarrow d=\tfrac{600}{9}\approx 66.67\). Monthly deposits: 200, 266.67, 333.33, 400, 466.67, 533.33, 600, 666.67, 733.33, 800. Sum = ₹5000.

Assertion–Reason Questions

Assertion (A): The sum of first \(n\) natural numbers is \(\tfrac{n(n+1)}{2}\).
Reason (R): It is a special case of \(S_n = \tfrac{n}{2}(a+l)\) with \(a=1,\ l=n\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a).

Assertion (A): If \(S_n = 3n^2+2n\), then \(a_n = 6n-1\).
Reason (R): \(a_n = S_n - S_{n-1}\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a). \(S_{n-1}=3(n-1)^2+2(n-1)=3n^2-4n+1\). \(a_n=S_n-S_{n-1}=6n-1\) ✓.

Frequently Asked Questions — Arithmetic Progressions

What is Part 2 — Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool in NCERT Class 10 Mathematics?

Part 2 — Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool is a key concept covered in NCERT Class 10 Mathematics, Chapter 5: Arithmetic Progressions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Part 2 — Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool step by step?

To solve problems on Part 2 — Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 5: Arithmetic Progressions?

The essential formulas of Chapter 5 (Arithmetic Progressions) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Part 2 — Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool important for the Class 10 board exam?

Part 2 — Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Part 2 — Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool?

Common mistakes in Part 2 — Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Part 2 — Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool?

End-of-chapter NCERT exercises for Part 2 — Sum of First n Terms of an AP | Class 10 Maths Ch 5 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 5, and solve at least one previous-year board paper to consolidate your understanding.

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