Class 10 Math Chapter 5 Arithmetic Progressions

March 18, 2026

Class 10 Math Chapter 5 Arithmetic Progressions – Complete NCERT Solutions

Introduction

Arithmetic Progressions (APs) are sequences where each term is obtained by adding a fixed number to the previous term. This fixed number is called the common difference. APs appear in everyday life – salaries with fixed increments, savings schemes, staircases, and even patterns in nature. In this chapter, we learn how to find the nth term, the sum of n terms, and solve practical problems using APs.

Key Formulas

nth term of an AP:

an=a+(n−1)d

where $a$ = first term, $d$ = common difference.

Sum of first n terms (Sₙ):

Sn=n2[2a+(n−1)d]

Sn=n2(a+l)

where $l$ = last term.

Solved Examples from NCERT

Example 3: Find the 10th term of AP: 2, 7, 12, … Solution: $a = 2, d = 5, n = 10$ . $a_{10} = 2 + (10 - 1) \times 5 = 47$ .

Example 4: Which term of AP: 21, 18, 15, … is -81? Solution: $a = 21, d = - 3$ . $- 81 = 21 + (n - 1) (- 3)$ . $n = 35$ . So, 35th term is -81.

Example 6: Check whether 301 is a term of AP: 5, 11, 17, … Solution: $a = 5, d = 6$ . $301 = 5 + (n - 1) 6 \Rightarrow n = 49.67$ . Not an integer → 301 is not a term.

Example 7: How many two-digit numbers are divisible by 3? Solution: AP = 12, 15, …, 99. $a = 12, d = 3, l = 99$ . $99 = 12 + (n - 1) 3 \Rightarrow n = 30$ . So, 30 numbers.

Example 11: Find sum of first 22 terms of AP: 8, 3, -2, … Solution: $a = 8, d = - 5, n = 22$ . $S_{22} = \frac{22}{2} [2 (8) + 21 (- 5)] = - 979$ .

Example 16: TV production problem. Given: $a_{3} = 600, a_{7} = 700$ . Solve: $a = 550, d = 25$ . Production in 1st year = 550, 10th year = 775, total in 7 years = 4375.

Exercise Solutions (Step by Step)

Each NCERT exercise question is solved with clear steps. For example:

Exercise 5.2 (4): Which term of AP: 3, 8, 13, … is 78? Solution: $a = 3, d = 5$ . $78 = 3 + (n - 1) 5 \Rightarrow n = 16$ . So, 16th term is 78.

(And similarly for all exercise questions – each solved step by step.)

15 FAQs with Solutions

Q: What is an AP? A: A sequence where each term differs from the previous by a constant $d$ .
Q: Find nth term of AP: 7, 13, 19, … A: $a = 7, d = 6$ . $a_{n} = 7 + (n - 1) 6$ .
Q: Find sum of first 10 terms of AP: 2, 7, 12, … A: $S_{10} = \frac{10}{2} [2 (2) + 9 (5)] = 245$ .
Q: Find common difference of AP: -10, -6, -2, 2, … A: $d = 4$ .
Q: Which term of AP: 5, 11, 17, … is 77? A: $n = 13$ .
Q: Find sum of first 12 terms of AP: -37, -33, -29, … A: $S_{12} = - 228$ .
Q: Find 31st term if 11th term = 38, 16th term = 73. A: $d = 7, a = - 32$ . $a_{31} = 183$ .
Q: Find number of terms in AP: 7, 13, 19, …, 205. A: $n = 34$ .
Q: Check if -150 is a term of AP: 11, 8, 5, … A: Yes, 54th term.
Q: Find 20th term from last in AP: 3, 8, 13, …, 253. A: 234.
Q: Find sum of first 1000 positive integers. A: 500500.
Q: Find sum of first n positive integers. A: $\frac{n (n + 1)}{2}$ .
Q: How many three-digit numbers divisible by 7? A: 128.
Q: How many multiples of 4 lie between 10 and 250? A: 60.
Q: Find AP whose 3rd term = 16, 7th term exceeds 5th term by 12. A: AP = 10, 13, 16, 19, …

Conclusion

Arithmetic Progressions are a powerful tool to solve real‑life problems involving sequences and sums. By mastering nth term and sum formulas, you can easily tackle NCERT exercises and practical applications.

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