Class 10 Maths Chapter 5 – Arithmetic Progressions | NCERT Solutions, Formulas & CBSE Board Questions (2025–26)

 Class 10 CBSE Maths – Chapter 5: Arithmetic Progressions

---

🔹 Key Concepts and Formulas

✅ What is an Arithmetic Progression (AP)?

An AP is a list of numbers in which the difference between any two consecutive terms is constant.

Let the first term be $a$ and the common difference be $d$.
Then the AP is: $a, a + d, a + 2d, a + 3d, \ldots$

✅ Important Formulas

1. General Term (n-th term) of an AP:
   $a_n = a + (n - 1)d$
   Where:

• $a$ = First term

• $d$ = Common difference

• $n$ = Term number

2. Sum of first n terms of an AP:
   $S_n = \frac{n}{2} [2a + (n - 1)d] \quad \text{or} \quad S_n = \frac{n}{2} (a + l)$
   Where:

• $a$ = First term

• $l$ = Last term = $a + (n - 1)d$

• $n$ = Number of terms

---

🔹 📘 NCERT Solved Examples (Highlights)

Example 1:

Find the 10th term of the AP: 2, 7, 12, 17, ...
$a = 2, d = 5, n = 10 \Rightarrow a_n = a + (n - 1)d = 2 + 9 × 5 = 47$

Example 2:

Find the sum of first 15 terms of AP: 4, 9, 14, ...
$a = 4, d = 5, n = 15 \Rightarrow S_n = \frac{15}{2}[2 × 4 + 14 × 5] = \frac{15}{2}[8 + 70] = \frac{15}{2} × 78 = 585$

---

🔹 📄 Exercise 5.1 – Identify AP and Find Terms

Q1. Identify which of the following are APs:

(i) 2, 4, 8, 16, ... ⛔ No (difference not constant)
(ii) -3, -5, -7, -9,... ✅ Yes, $d = -2$
(iii) 3, 3, 3, 3,... ✅ Yes, $d = 0$

Q2. Find the 15th term of AP: 7, 10, 13, ...

$a = 7, d = 3, n = 15 \Rightarrow a_{15} = 7 + 14 × 3 = 49$

---

🔹 📄 Exercise 5.2 – General Term Formula

Q1. Which term of the AP: 3, 8, 13, 18, ... is 78?

$a = 3, d = 5 \Rightarrow a_n = 3 + (n - 1)5 = 78$
$5n - 2 = 78 \Rightarrow n = 16 \Rightarrow 78 ext{ is the 16th term}$

Q2. Find the number of terms in AP: 7, 13, 19, ..., 205

$a = 7, d = 6, l = 205 \Rightarrow 205 = a + (n - 1)d \Rightarrow 205 = 7 + (n - 1)6 \Rightarrow n = 34$

---

🔹 📄 Exercise 5.3 – Sum of First n Terms

Q1. Find the sum of first 25 terms of AP: 3, 6, 9, ...

$a = 3, d = 3, n = 25 \Rightarrow S_n = \frac{25}{2}[2 × 3 + 24 × 3] = \frac{25}{2}[6 + 72] = 975$

Q2. If the 3rd and 7th terms of an AP are 4 and -8, find the sum of first 20 terms.

$a + 2d = 4, a + 6d = -8 \Rightarrow d = -3, a = 10$
$S_{20} = \frac{20}{2}[2 × 10 + 19 × (-3)] = 10[20 - 57] = -370$

---

🔹 Summary and Revision Notes

• APs have constant difference between terms (d)

• nth term tells you any position value: $a_n = a + (n - 1)d$

• Sum formula helps calculate total value quickly

• Solving AP problems often involves rearranging the nth term or sum formulas

---

🔹 CBSE Previous Year Questions – Arithmetic Progressions

✅ 1-Mark Questions

(2020) Find the 11th term of AP: 2, 7, 12, ...
$a = 2, d = 5, a_{11} = 2 + 10 × 5 = 52$

✅ 2-Mark Questions

(2021) Which term of AP: 5, 11, 17, ... is 95?
$a = 5, d = 6 \Rightarrow 95 = 5 + (n - 1)6 \Rightarrow n = 16$

✅ 3-Mark Questions

(2019) Find number of terms in AP: 7, 10.5, 14, ..., 84
$a = 7, d = 3.5 \Rightarrow 84 = 7 + (n - 1)3.5 \Rightarrow n = 23$

(2018) The 4th term of an AP is 0. The 9th term is 20. Find the first term and common difference.
$a + 3d = 0, a + 8d = 20 \Rightarrow d = 4, a = -12$

✅ 4-Mark Word Problems

(2020) Sum of 6 terms of an AP is 168. The 1st term is 11. Find the common difference.
$S_6 = \frac{6}{2}[2 × 11 + 5d] = 168 \Rightarrow d = 7$

(2017) The sum of first 7 terms of an AP is 182. If the fourth term is 29, find the AP.
$a + 3d = 29, S_7 = \frac{7}{2}[2a + 6d] = 182 \Rightarrow a = 10, d = 3$
✅ AP: 10, 13, 16, ...