NCERT Class 10 Maths Chapter 5

NCERT Class 10 Maths Chapter 5 – Arithmetic Progressions (AP) Simplified Notes & Tricks

NCERT-class-10-maths-chapter-5: Mathematics can be tricky, but with the right approach and practice, it becomes easy and enjoyable. We learn about sequences of numbers and how to find missing terms, sums, and general formulas. This guide simplifies key concepts, provides shortcut tricks, and includes multiple-choice questions (MCQs) with solutions.

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What is an Arithmetic Progression (AP)?

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant.

Example:

2, 5, 8, 11, 14, … is an AP where the common difference (d) = 5 – 2 = 3.

General Form of an AP:

A general AP looks like this:
a, a + d, a + 2d, a + 3d, …

Where:

• a = First term
• d = Common difference
• n = Number of terms
• l = Last term

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Important Formulas of Arithmetic Progressions

1️⃣ nth Term (General Term) of an AP

an=a+(n−1)dan​=a+(n−1)d

✔️ Use Case: Find any term of an AP without writing all terms.

🔹 Example:
Find the 10th term of the AP: 3, 7, 11, 15, …

🔹 Solution:

• a = 3
• d = 7 – 3 = 4
• n = 10

Using the formula:

an=3+(10−1)×4=3+9×4=3+36=39an​=3+(10−1)×4=3+9×4=3+36=39

✅ Answer: 39

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2️⃣ Sum of First n Terms of an AP

Sn=n2[2a+(n−1)d]

Sn​=2n​[2a+(n−1)d]

or

Sn=n2(a+l)

Sn​=2n​(a+l)

✔️ Use Case: Find the sum of a given number of terms in an AP.

🔹 Example:
Find the sum of the first 15 terms of the AP: 2, 5, 8, 11, …

🔹 Solution:

• a = 2
• d = 5 – 2 = 3
• n = 15

Using the formula:

S15=152[2×2+(15−1)×3]S15​=215​[2×2+(15−1)×3]=152[4+42]=152×46=6902=345=215​[4+42]=215​×46=2690​=345

✅ Answer: 345

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Finding the Common Difference

d=a2−a1d=a2​−a1​

✔️ Use Case: Find the difference between consecutive terms in an AP.

🔹 Example:
If the 3rd term of an AP is 10 and the 7th term is 26, find d.

🔹 Solution:
Using the nth term formula:

a3=a+2d=10a3​=a+2d=10a7=a+6d=26a7​=a+6d=26

Subtracting equations:

(a+6d)−(a+2d)=26−10

$$(a+6d)-(a+2d)=26-10$$

$$4$$

$$d=16$$

d=164=4d=416​=4

✅ Answer: 4

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MCQs with Solutions

Q1. Which of the following sequences is an AP?

(a) 2, 5, 8, 11, …
(b) 3, 6, 10, 15, …
(c) 1, 4, 9, 16, …
(d) 5, 10, 20, 40, …

🔹 Solution:
Checking the common difference d:
✔️ For (a): $d=5-2=3,8-5=3$ ✅ AP
✔️ For (b): $d=6-3=3,10-6=4$ ❌ Not AP
✔️ For (c): $1,4,9,16$ (Squares) ❌ Not AP
✔️ For (d): $5,10,20,40$ (Multiplication) ❌ Not AP

✅ Correct Answer: (a) 2, 5, 8, 11, …

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Find the 12th term of the AP: 4, 9, 14, 19, …

(a) 54
(b) 59
(c) 64
(d) 69

🔹 Solution:

• a = 4
• d = 9 – 4 = 5
• n = 12

Using the formula:

an=4+(12−1)×5=4+55=59an​=4+(12−1)×5=4+55=59

✅ Correct Answer: (b) 59

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Q3. The sum of the first 20 terms of the AP: 5, 10, 15, 20, … is:

(a) 950
(b) 1000
(c) 1050
(d) 1100

🔹 Solution:

• a = 5
• d = 10 – 5 = 5
• n = 20

Using the sum formula:

Sn=202[2×5+(20−1)×5]Sn​=220​[2×5+(20−1)×5]$$=10[10+95]=10\times 105=1050$$

✅ Correct Answer: (c) 1050

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Tricks to Solve AP Problems Faster

✔️ Shortcut for nth Term: If given the first term (a) and the last term (l), use:

an=l−(n−1)dan​=l−(n−1)d

✔️ Quickly Find Sum of n Terms:
If l is known, use

Sn=n2(a+l)Sn​=2n​(a+l)

✔️ Identify AP Easily: The difference between consecutive terms should be the same.

To help you master Arithmetic Progressions (AP), here are some additional problems categorized into different difficulty levels. Each question includes a step-by-step solution.

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Basic Level Questions

Q1. Find the 15th term of the AP: 7, 13, 19, 25, …

(a) 91
(b) 85
(c) 95
(d) 89

🔹 Solution:

• a = 7
• d = 13 – 7 = 6
• n = 15

Using the formula for the nth term:

an=a+(n−1)da_n = a + (n-1) dan​=a+(n−1)da15=7+(15−1)×6a_{15} = 7 + (15-1) \times 6a15​=7+(15−1)×6$$=7+14\times 6$$$$=7+84=91$$

✅ Correct Answer: (a) 91

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Q2. Which of the following sequences is an AP?

(a) 3, 6, 12, 24, …
(b) 10, 20, 30, 40, …
(c) 1, 4, 9, 16, …
(d) 5, 15, 25, 40, …

🔹 Solution:

• (a): Common difference changes ($6-3=3$, $12-6=6$), ❌ Not AP
• (b): Common difference = $20-10=10$, $30-20=10$, ✅ AP
• (c): Square series (12,22,321^2, 2^2, 3^212,22,32), ❌ Not AP
• (d): Common difference changes ($15-5=10$, $25-15=10$, but $40-25=15$), ❌ Not AP

✅ Correct Answer: (b) 10, 20, 30, 40, …

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Intermediate Level Questions

Q3. The 8th term of an AP is 23, and the 12th term is 35. Find the first term and the common difference.

🔹 Solution:
Using the nth term formula:

an=a+(n−1)da_n = a + (n-1) dan​=a+(n−1)d

For 8th term (a₈ = 23):

$$a+7d=23$$

For 12th term (a₁₂ = 35):

$$a+11d=35$$

Subtracting equations:

$$(a+11d)-(a+7d)=35-23$$$$4d=12$$$$d=3$$

Substituting d = 3 in a + 7d = 23:

$$a+7(3)=23$$$$a+21=23$$$$a=2$$

✅ Answer: First term = 2, Common difference = 3

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Q4. Find the sum of the first 25 terms of the AP: 4, 9, 14, 19, …

🔹 Solution:
Given:

• a = 4
• d = 9 – 4 = 5
• n = 25

Using the sum formula:

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} [2a + (n-1) d]Sn​=2n​[2a+(n−1)d]S25=252[2(4)+(25−1)×5]S_{25} = \frac{25}{2} [2(4) + (25-1) \times 5]S25​=225​[2(4)+(25−1)×5]=252[8+120]= \frac{25}{2} [8 + 120]=225​[8+120]=252×128= \frac{25}{2} \times 128=225​×128=32002=1600= \frac{3200}{2} = 1600=23200​=1600

✅ Answer: 1600

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Q5. The 5th term of an AP is 20, and the 15th term is 50. Find the sum of the first 20 terms.

🔹 Solution:
Using the nth term formula:

an=a+(n−1)da_n = a + (n-1) dan​=a+(n−1)d

For 5th term (a₅ = 20):

$$a+4d=20$$

For 15th term (a₁₅ = 50):

$$a+14d=50$$

Subtracting equations:

$$(a+14d)-(a+4d)=50-20$$$$10d=30$$$$d=3$$

Substituting d = 3 in a + 4d = 20:

$$a+4(3)=20$$$$a+12=20$$$$a=8$$

Now, find S₁₀:

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} [2a + (n-1) d]Sn​=2n​[2a+(n−1)d]S20=202[2(8)+(20−1)×3]S_{20} = \frac{20}{2} [2(8) + (20-1) \times 3]S20​=220​[2(8)+(20−1)×3]$$=10[16+57]$$$$=10\times 73=730$$

✅ Answer: Sum of first 20 terms = 730

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Advanced Level Questions

Q6. The sum of the first 10 terms of an AP is 260, and the sum of the first 15 terms is 555. Find the first term and common difference.

🔹 Solution:
We have:

S10=260S_{10} = 260S10​=260S15=555S_{15} = 555S15​=555

Using sum formula:

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} [2a + (n-1) d]Sn​=2n​[2a+(n−1)d]

For S₁₀ = 260:

102[2a+9d]=260\frac{10}{2} [2a + 9d] = 260210​[2a+9d]=260$$5[2a+9d]=260$$$$2a+9d=52$$

For S₁₅ = 555:

152[2a+14d]=555\frac{15}{2} [2a + 14d] = 555215​[2a+14d]=555$$7.5[2a+14d]=555$$$$2a+14d=74$$

Solving:

$$(2a+14d)-(2a+9d)=74-52$$$$5d=22$$$$d=4.4$$

Substituting in 2a + 9(4.4) = 52:

$$2a+39.6=52$$$$2a=12.4$$$$a=6.2$$

✅ Answer: First term a = 6.2, Common difference d = 4.4

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