Chapter 5 : Arithmetic Progressions Class 10 Maths NCERT Solutions

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Contents

Class 10 Maths NCERT Solutions - Chapter wise

1

Chapter 1: Real Numbers

2

Chapter 2: Polynomials

3

Chapter 3 : Pair of Linear Equations in Two Variables

4

Chapter 4 : Quadratic Equations

5

Chapter 5 : Arithmetic Progressions

6

Chapter 6 : Triangles

7

Chapter 7 : Coordinate Geometry

8

Chapter 8: Introduction to Trigonometry

9

Chapter 9: Some Applications of Trigonometry

10

Chapter 10 : Circles

11

Chapter 11 : Areas Related to Circles

12

Chapter 12: Surface Areas and Volumes

13

Chapter 13: Statistics

14

Chapter 14 : Probability

Chapter Notes

Chapter 5: Arithmetic Progressions — Exercise 5.1

Ever noticed how the rungs of a ladder or the steps of a staircase are at a fixed distance? That’s an Arithmetic Progression (AP) in real life—mastering this helps you predict patterns in everything from savings to physics!

Q1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

  • The Logic: For a sequence to be an AP, the difference between consecutive terms must be constant (the same). Here, we are adding a fixed amount for every extra km.

  • The Steps: * Fare for 1st km (a) = 15

    • Fare for 2nd km = 15 + 8 = 23

    • Fare for 3rd km = 23 + 8 = 31

    • The sequence is 15, 23, 31… Here, the difference (d) is always 8.

  • The Result: Yes, it forms an AP because the common difference is constant (8).

ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

  • The Logic: In an AP, we add or subtract a fixed value, not a percentage or fraction of the remaining amount.

  • The Steps: * Let initial volume be V.

    • Next volume = V – (1/4)V = 3/4 V

    • Next volume = 3/4 V – 1/4(3/4 V) = 9/16 V

    • The differences are (3/4 V – V) and (9/16 V – 3/4 V). These are not equal!

  • The Result: No, it does not form an AP because the common difference is not constant.

Q2. Write first four terms of the AP, when the first term a and the common difference d are given as follows:

i) a = 10, d = 10

  • The Logic: Start with ‘a’ and keep adding ‘d’ to get the next term.

  • The Steps: * 1st term = 10

    • 2nd term = 10 + 10 = 20

    • 3rd term = 20 + 10 = 30

    • 4th term = 30 + 10 = 40

  • The Result: The terms are 10, 20, 30, 40.

ii) a = -2, d = 0

  • Human Tip: If d = 0, the sequence never changes! It’s like walking in place.

  • The Steps: -2 + 0 = -2…

  • The Result: The terms are -2, -2, -2, -2.

Q3. For the following APs, write the first term and the common difference:

i) 3, 1, -1, -3, …

  • The Logic: The first number you see is ‘a’. To find ‘d’, subtract the 1st term from the 2nd term (Term 2 – Term 1).

  • The Steps: * First term (a) = 3

    • Common difference (d) = 1 – 3 = -2

  • The Result: a = 3, d = -2

ii) 1/3, 5/3, 9/3, 13/3, …

  • The Steps: * a = 1/3

    • d = 5/3 – 1/3 = 4/3

  • The Result: a = 1/3, d = 4/3

Q4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

i) 2, 4, 8, 16, …

  • The Logic: Check if d is the same everywhere.

  • The Steps: * 4 – 2 = 2

    • 8 – 4 = 4

    • The differences are 2 and 4. They are not the same!

  • The Result: This is not an AP.

ii) 2, 5/2, 3, 7/2, …

  • The Logic: Convert to decimals if it’s easier to see: 2, 2.5, 3, 3.5…

  • The Steps: * d = 2.5 – 2 = 0.5 (or 1/2).

    • Since d is constant, it is an AP.

    • Next three terms: 3.5 + 0.5 = 4; 4 + 0.5 = 4.5 (9/2); 4.5 + 0.5 = 5.

  • The Result: It is an AP. d = 1/2. Next terms: 4, 9/2, 5.

Summary Table

Concept Key Secret Formula / Rule
First Term (a) The starting point The first number in the list
Common Difference (d) The “Step” size d = (any term) – (previous term)
Condition for AP Must be steady ‘d’ must be the same throughout
Finding next terms Keep building a_n = a_{n-1} + d

Chapter 5: Arithmetic Progressions — Exercise 5.2

Ever wondered how to predict the 100th step of a pattern without counting every single one? This exercise gives you the “Time Machine” formula of Math to find any specific term in a sequence instantly!

The Master Formula

Before we dive in, remember this one formula. It is the heart of this entire exercise:

an = a + (n – 1)d

  • an: The term you are looking for (The nth term).

  • a: The very first number in the list.

  • n: The position of the term (e.g., 10th, 50th).

  • d: The “Common Difference” (What you add each time).

Q1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:

(i) a = 7, d = 3, n = 8, find an

  • The Logic: We have the starting point, the gap, and the position. We just need to plug them into our “Time Machine” formula.

  • The Steps:

    • an = a + (n – 1)d

    • an = 7 + (8 – 1)3

    • an = 7 + (7)3

    • an = 7 + 21 = 28

  • The Result: an = 28

(ii) a = -18, n = 10, an = 0, find d

  • The Logic: Here, we know where we started and where we ended up at the 10th step. We need to find the “gap” (d) between each step.

  • The Steps:

    • 0 = -18 + (10 – 1)d

    • 18 = 9d

    • d = 18 / 9 = 2

  • The Result: d = 2

Q2. Choose the correct choice in the following and justify:

(i) 30th term of the AP: 10, 7, 4, … is:

  • The Logic: First, identify our variables. a = 10. To find ‘d’, subtract the first from the second (7 – 10 = -3). We want the 30th term, so n = 30.

  • The Steps:

    • a30 = 10 + (30 – 1)(-3)

    • a30 = 10 + (29)(-3)

    • a30 = 10 – 87 = -77

  • The Result: -77 (Option C)

Q4. Which term of the AP: 3, 8, 13, 18, … is 78?

  • The Logic: Here, we know the value (78) but we don’t know the “rank” or position (n). We are solving for n.

  • Human Tip: Since ‘n’ represents a position in a line, it must always be a positive whole number. If you get a fraction, double-check your math!

  • The Steps:

    • a = 3, d = 8 – 3 = 5, an = 78

    • 78 = 3 + (n – 1)5

    • 75 = (n – 1)5

    • 15 = n – 1

    • n = 16

  • The Result: 78 is the 16th term.

Q7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

  • The Logic: This is a classic board exam question! We have two clues. We’ll create two equations and solve them like a puzzle to find ‘a’ and ‘d’.

  • The Steps:

    1. Equation 1 (11th term): a + 10d = 38

    2. Equation 2 (16th term): a + 15d = 73

    3. Subtract Eq 1 from Eq 2: (15d – 10d) = 73 – 38 => 5d = 35 => d = 7

    4. Find ‘a’: a + 10(7) = 38 => a + 70 = 38 => a = -32

    5. Now find 31st term: a31 = -32 + (31 – 1)7

    6. a31 = -32 + (30)7 = -32 + 210 = 178

  • The Result: The 31st term is 178.

Q13. How many three-digit numbers are divisible by 7?

  • Pro-Tip: The first 3-digit number is 100, but is it divisible by 7? No. The first one is 105. The last one before 1000 is 994.

  • The Steps:

    • AP: 105, 112, 119, … , 994

    • a = 105, d = 7, an = 994

    • 994 = 105 + (n – 1)7

    • 889 = (n – 1)7

    • 127 = n – 1

    • n = 128

  • The Result: There are 128 such numbers.

Student Summary Table (Cheat Sheet)

If you want to find… Use this Logic Key Formula
Any specific term Use the “Time Machine” an = a + (n – 1)d
The Gap (d) Subtract backwards d = a2 – a1
The Position (n) Solve for n n = [(an – a) / d] + 1
Two Unknowns Use the two-equation method Simultaneous Equations

Chapter 5: Arithmetic Progressions — Exercise 5.3

Ever wondered how to quickly add up a long list of numbers, like your savings over a year or seats in a stadium? This exercise teaches you the “Sum Formula,” a superpower that lets you calculate totals in seconds without adding them one by one!

Q1. Find the sum of the following APs:

i) 2, 7, 12, …, to 10 terms

  • The Logic: When we need the sum of a sequence, we use the Sum of n terms (Sn) formula. We just need to identify the first term (a), the common difference (d), and how many terms we are adding (n).

  • The Steps:

    • a = 2, d = 7 – 2 = 5, n = 10

    • Formula: Sn = n/2 [2a + (n – 1)d]

    • S10 = 10/2 [2(2) + (10 – 1)5]

    • S10 = 5 [4 + (9 x 5)]

    • S10 = 5 [4 + 45] = 5 x 49

  • The Result: The sum is 245.

Q2. Find the sums given below:

i) 7 + 10.5 + 14 + … + 84

  • The Logic: Here, we know the last term (l = 84), but we don’t know “n” (how many numbers are there). So, first we find “n” using the an formula, then we find the sum.

  • The Steps:

    • a = 7, d = 3.5, an = 84

    • First, find n: 84 = 7 + (n – 1)3.5

    • 77 = (n – 1)3.5 => n – 1 = 77 / 3.5 = 22 => n = 23

    • Now use the Short Formula: Sn = n/2 (a + l)

    • S23 = 23/2 (7 + 84) = 23/2 (91) = 2093 / 2

  • The Result: The sum is 1046.5.

Q3. In an AP:

i) Given a = 5, d = 3, an = 50, find n and Sn.

  • The Logic: This is a classic “missing piece” puzzle. Use the nth term formula to find n first.

  • The Steps:

    • 50 = 5 + (n – 1)3

    • 45 = (n – 1)3 => n – 1 = 15 => n = 16

    • Sn = 16/2 (5 + 50) = 8 x 55

  • The Result: n = 16, Sn = 440.

Q4. How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?

  • The Logic: This involves a Quadratic Equation. We know the total sum, but we need to solve for ‘n’.

  • The Steps:

    • a = 9, d = 8, Sn = 636

    • 636 = n/2 [2(9) + (n – 1)8]

    • 636 = n/2 [18 + 8n – 8]

    • 636 = n [9 + 4n – 4] => 636 = n [4n + 5]

    • 4n^2 + 5n – 636 = 0

    • Solving the quadratic gives n = 12 (ignoring the negative value).

  • The Result: 12 terms must be taken.

Q7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

  • Pro-Tip: If you have the last term (a22), find ‘a’ first using the an formula, then use the simple sum formula!

  • The Steps:

    • a22 = a + 21d => 149 = a + 21(7)

    • 149 = a + 147 => a = 2

    • S22 = 22/2 (a + a22) = 11 (2 + 149) = 11 x 151

  • The Result: The sum is 1661.

Q9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

  • Human Tip: Notice the pattern? 7^2 = 49 and 17^2 = 289. This is a huge hint that the answer will involve n^2!

  • The Steps:

    • S7 = 7/2 [2a + 6d] = 49 => a + 3d = 7

    • S17 = 17/2 [2a + 16d] = 289 => a + 8d = 17

    • Subtracting gives 5d = 10 => d = 2. Then a = 1.

    • Sn = n/2 [2(1) + (n – 1)2] = n/2 [2n] = n^2

  • The Result: The sum of n terms is n^2.

Summary Table 

Goal Formula to Use When to use it?
Sum (General) Sn = n/2 [2a + (n – 1)d] When you know a, d, and n.
Sum (Short) Sn = n/2 (a + l) When you know the first and last term.
Find n an = a + (n – 1)d When the number of terms is missing.
Sum of First n Odd Nos Sn = n^2 Quick trick for odd number sequences.

Chapter 5: Arithmetic Progressions — Exercise 5.4 (Optional)

Think of this exercise as the “Boss Level” of AP! While it is marked as optional, these questions are the ones that truly test your logic and help you stand out as a topper in competitive exams.

Q1. Which term of the AP: 121, 117, 113, … , is its first negative term?

  • The Logic: The numbers are decreasing, so eventually, they will drop below zero. We need to find the first position (n) where the term (an) becomes less than 0.

  • The Steps:

    • Here, first term a = 121

    • Common difference d = 117 – 121 = -4

    • We want to find ‘n’ such that an < 0

    • Formula: a + (n – 1)d < 0

    • 121 + (n – 1)(-4) < 0

    • 121 – 4n + 4 < 0

    • 125 – 4n < 0

    • 125 < 4n

    • n > 125 / 4

    • n > 31.25

  • Human Tip: Since ‘n’ must be a whole number (you can’t have a 31.25th term!), the next whole number after 31.25 is 32.

  • The Result: The 32nd term is the first negative term.

Q2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

  • The Logic: We have two conditions which will give us two equations. Solving them will reveal ‘a’ and ‘d’. Remember, ‘d’ can be positive or negative, so we might have two possible answers!

  • The Steps:

    • Condition 1 (Sum): (a + 2d) + (a + 6d) = 6 => 2a + 8d = 6 => a + 4d = 3

    • Condition 2 (Product): (a + 2d)(a + 6d) = 8

    • From Condition 1, a = 3 – 4d. Substitute this into the product:

    • (3 – 4d + 2d)(3 – 4d + 6d) = 8

    • (3 – 2d)(3 + 2d) = 8

    • 9 – 4d^2 = 8 (Using the (x-y)(x+y) identity)

    • 1 = 4d^2 => d^2 = 1/4 => d = 1/2 or -1/2

    • Case 1: If d = 1/2, then a = 1. Sum of 16 terms (S16) = (16/2)[2(1) + 15(1/2)] = 8[2 + 7.5] = 76.

    • Case 2: If d = -1/2, then a = 5. S16 = (16/2)[2(5) + 15(-1/2)] = 8[10 – 7.5] = 20.

  • The Result: The sum of the first sixteen terms is either 76 or 20.

Q3. A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2.5 m apart, what is the length of the wood required for the rungs?

  • The Logic: This is a real-life AP problem. The rungs represent terms in an AP. To find the “total wood,” we need to find the Sum (Sn). First, we must calculate how many rungs (n) there are.

  • The Steps:

    • Total height of ladder = 2.5 m = 250 cm.

    • Distance between rungs = 25 cm.

    • Number of rungs (n) = (Total length / Distance) + 1 = (250 / 25) + 1 = 11 rungs.

    • Length of first rung (a) = 45 cm.

    • Length of last rung (l) = 25 cm.

    • Pro-Tip: When you know the first and last term, use the simpler sum formula: Sn = (n/2)(a + l).

    • S11 = (11/2)(45 + 25) = (11/2)(70) = 11 x 35 = 385.

  • The Result: The total length of wood required is 385 cm.

Q4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

  • The Logic: This sounds like a puzzle! We need to find a house ‘x’ where:

    Sum of houses (1 to x-1) = Sum of houses (x+1 to 49).

     

  • The Steps:

     

    • Sum of (1 to x-1) = Sx-1

       

    • Sum of (x+1 to 49) = S49 – Sx

       

    • Equation: Sx-1 = S49 – Sx

    • [(x-1)/2][1 + (x-1)] = [49/2][1 + 49] – [x/2][1 + x]

    • [(x-1)x] / 2 = (49 x 50) / 2 – [x(x+1)] / 2

    • Multiply whole equation by 2 to clear denominators: x^2 – x = 2450 – x^2 – x

    • 2x^2 = 2450

    • x^2 = 1225

    • x = 35

  • The Result: The value of x is 35.

Student Summary Table (Arithmetic Progressions)

Formula Name When to Use? Standard Format
n-th Term To find a specific house/rung number an = a + (n – 1)d
Sum (Standard) When you have a, d, and n Sn = (n/2)[2a + (n – 1)d]
Sum (Last Term) When you know the end point (l) Sn = (n/2)(a + l)
Common Difference To check if a sequence is an AP d = a2 – a1

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