Class 10 Maths Chapter 5 Arithmetic Progressions

Class 10 Maths Chapter 5 Arithmetic Progressions – Exercise 5.3 NCERT Solutions

Introduction

Exercise 5.3 focuses on finding the sum of the first n terms of an Arithmetic Progression (AP). This is useful in real‑life applications like calculating total costs, distances, or series sums. You’ll learn two formulas for the sum of AP and how to apply them in different problems.

Formula Used

Sum of n terms of AP (using nth term):

Sn=n2(a+an)

Sum of n terms of AP (using common difference):

Sn=n2(2a+(n−1)d)

Where:

$a$ = first term
$d$ = common difference
$n$ = number of terms
$a_{n}$ = nth term

NCERT Questions with Solutions (10)

Q1. Find the sum of first 10 terms of AP: 2, 7, 12, …

Solution: $a = 2, d = 5, n = 10$ . $S_{10} = \frac{10}{2} [2 (2) + (10 - 1) (5)] = 5 [4 + 45] = 245$ .

Q2. Find the sum of first 20 terms of AP: 1, 3, 5, …

Solution: $a = 1, d = 2, n = 20$ . $S_{20} = \frac{20}{2} [2 (1) + (19) (2)] = 10 [2 + 38] = 400$ .

Q3. Find the sum of first 15 terms of AP: 5, 10, 15, …

Solution: $a = 5, d = 5, n = 15$ . $S_{15} = \frac{15}{2} [2 (5) + (14) (5)] = \frac{15}{2} [10 + 70] = \frac{15}{2} \cdot 80 = 600$ .

Q4. Find the sum of first 12 terms of AP: -3, -1, 1, …

Solution: $a = - 3, d = 2, n = 12$ . $S_{12} = \frac{12}{2} [2 (- 3) + (11) (2)] = 6 [- 6 + 22] = 96$ .

Q5. Find the sum of first 10 terms of AP: 10, 7, 4, …

Solution: $a = 10, d = - 3, n = 10$ . $S_{10} = \frac{10}{2} [20 + (9) (- 3)] = 5 [20 - 27] = - 35$ .

Q6. Find the sum of first 30 terms of AP: 3, 6, 9, …

Solution: $a = 3, d = 3, n = 30$ . $S_{30} = \frac{30}{2} [6 + (29) (3)] = 15 [6 + 87] = 15 \cdot 93 = 1395$ .

Q7. Find the sum of first 18 terms of AP: 8, 14, 20, …

Solution: $a = 8, d = 6, n = 18$ . $S_{18} = \frac{18}{2} [16 + (17) (6)] = 9 [16 + 102] = 9 \cdot 118 = 1062$ .

Q8. Find the sum of first 50 terms of AP: 1, 4, 7, …

Solution: $a = 1, d = 3, n = 50$ . $S_{50} = \frac{50}{2} [2 + (49) (3)] = 25 [2 + 147] = 25 \cdot 149 = 3725$ .

Q9. Find the sum of first 22 terms of AP: 100, 97, 94, …

Solution: $a = 100, d = - 3, n = 22$ . $S_{22} = \frac{22}{2} [200 + (21) (- 3)] = 11 [200 - 63] = 11 \cdot 137 = 1507$ .

Q10. Find the sum of first 40 terms of AP: 0, 10, 20, …

Solution: $a = 0, d = 10, n = 40$ . $S_{40} = \frac{40}{2} [0 + (39) (10)] = 20 \cdot 390 = 7800$ .

FAQs (10 from NCERT)

Q: What is the formula for sum of n terms of AP? A: $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .
Q: What is the nth term formula? A: $a_{n} = a + (n - 1) d$ .
Q: Can sum of AP be negative? A: Yes, if terms are negative.
Q: Can sum of AP be zero? A: Yes, if positive and negative terms cancel.
Q: What is the sum of first n natural numbers? A: $\frac{n (n + 1)}{2}$ .
Q: What is the sum of first n odd numbers? A: $n^{2}$ .
Q: What is the sum of first n even numbers? A: $n (n + 1)$ .
Q: Why is AP important? A: It models real‑life sequences like costs, distances, salaries.
Q: What is a finite AP sum? A: Sum of limited terms.
Q: Why is this exercise important? A: It builds skills for calculating series sums efficiently.

Conclusion

Exercise 5.3 has 10 solved questions and 10 FAQs that strengthen your understanding of the sum of n terms in arithmetic progressions. This builds the foundation for advanced AP applications in Class 10 Maths.

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