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Class 10 Maths Part 1 Chapter 3 Arithmetic Progression Practice Set 3.2 Solutions (Maharashtra Board)

Welcome to our comprehensive guide for Maharashtra Board Class 10 Maths Part 1. In this post, we provide complete, step-by-step Textbook Exercise Solutions for Chapter 3: Arithmetic Progression, specifically focusing on Practice Set 3.2. These verified solutions will help you master sequence calculations, score maximum marks in your board exams, and easily find the nth term of any sequence.

Question 1: Write the correct number in the given boxes from the following A.P.

  1. Given A.P.: \( 1, 8, 15, 22, \dots \)[cite: 1]

    Calculation:

    • Here \( a = 1 \), \( t_1 = 1 \), \( t_2 = 8 \), \( t_3 = 15 \)[cite: 2]
    • \( t_2 - t_1 = 8 - 1 = 7 \)[cite: 2]
    • \( t_3 - t_2 = 15 - 8 = 7 \)[cite: 2]

    Final Answer: \( d = 7 \)[cite: 2]

  2. Given A.P.: \( 3, 6, 9, 12, \dots \)[cite: 1]

    Calculation:

    • Here \( t_1 = 3 \), \( t_2 = 6 \), \( t_3 = 9 \), \( t_4 = 12 \)[cite: 2]
    • \( t_2 - t_1 = 6 - 3 = 3 \) (Note: Derived from sequence pattern)
    • \( t_3 - t_2 = 9 - 6 = 3 \)[cite: 2]

    Final Answer: \( d = 3 \)[cite: 2]

  3. Given A.P.: \( -3, -8, -13, -18, \dots \)[cite: 1]

    Calculation:

    • Here \( t_1 = -3 \), \( t_2 = -8 \), \( t_3 = -13 \), \( t_4 = -18 \)[cite: 2]
    • \( t_2 - t_1 = -8 - (-3) = -5 \)[cite: 2]
    • \( t_3 - t_2 = -13 - (-8) = -5 \)[cite: 2]

    Final Answer: \( d = -5 \)[cite: 2]

  4. Given A.P.: \( 70, 60, 50, 40, \dots \)[cite: 1]

    Calculation:

    • Here \( t_1 = 70 \), \( t_2 = 60 \), \( t_3 = 50 \)[cite: 2]
    • \( a = 70 \)[cite: 2]
    • \( t_2 - t_1 = 60 - 70 = -10 \)

    Final Answer: \( d = -10 \)[cite: 2]

Question 2: Decide whether following sequence is an A.P., if so find the 20th term of the progression.

Given Sequence: \( -12, -5, 2, 9, 16, 23, 30, \dots \)[cite: 2]

Calculation (Checking for A.P.):

  • Here, \( t_1 = -12 \), \( t_2 = -5 \), \( t_3 = 2 \), \( t_4 = 9 \)[cite: 2]
  • \( t_2 - t_1 = -5 - (-12) = -5 + 12 = 7 \)[cite: 2]
  • \( t_3 - t_2 = 2 - (-5) = 2 + 5 = 7 \)[cite: 2]
  • \( t_4 - t_3 = 9 - 2 = 7 \)[cite: 2]
  • \( t_2 - t_1 = t_3 - t_2 = \dots = 7 = d = \text{constant} \)[cite: 2]

The difference between two consecutive terms is constant. Therefore, the given sequence is an Arithmetic Progression (A.P.).[cite: 3]

Key Formula: \( t_n = a + (n-1)d \)[cite: 3]

Calculation (Finding 20th term):

  • Substitute \( a = -12 \), \( d = 7 \), and \( n = 20 \)[cite: 3]
  • \( t_{20} = -12 + (20 - 1)7 \)[cite: 3]
  • \( t_{20} = -12 + 19 \times 7 \)[cite: 3]
  • \( t_{20} = -12 + 133 \)[cite: 3]
  • \( t_{20} = 121 \)[cite: 3]

Final Answer: The 20th term of the given A.P. is 121.[cite: 3]

Question 3: Given Arithmetic Progression is 12, 16, 20, 24, ... Find the 24th term of this progression.

Given: A.P. is \( 12, 16, 20, 24, \dots \)[cite: 3]

  • First term \( a = 12 \)[cite: 3]
  • Common difference \( d = 16 - 12 = 4 \)[cite: 3]

Formula: \( t_n = a + (n-1)d \)[cite: 3]

Calculation:

  • \( t_{24} = 12 + (24 - 1)4 \)[cite: 3]
  • \( t_{24} = 12 + 23 \times 4 \)[cite: 3]
  • \( t_{24} = 12 + 92 \)[cite: 3]
  • \( t_{24} = 104 \)[cite: 3]

Final Answer: The 24th term of the given A.P. is 104.[cite: 3]

Question 4: Find the 19th term of the following A.P. 7, 13, 19, 25, ...

Given: A.P. is \( 7, 13, 19, 25, \dots \)[cite: 3]

  • First term \( a = 7 \)[cite: 3]
  • Common difference \( d = 13 - 7 = 6 \)[cite: 3]

Formula: \( t_n = a + (n-1)d \)[cite: 3]

Calculation:

  • \( t_{19} = 7 + (19 - 1)6 \)[cite: 3]
  • \( t_{19} = 7 + 18 \times 6 \)[cite: 3]
  • \( t_{19} = 7 + 108 \)[cite: 3]
  • \( t_{19} = 115 \)[cite: 4]

Final Answer: The 19th term of the given A.P. is 115.[cite: 4]

Question 5: Find the 27th term of the following A.P. 9, 4, -1, -6, -11, ...

Given: A.P. is \( 9, 4, -1, -6, -11, \dots \)[cite: 4]

  • First term \( a = 9 \)[cite: 4]
  • Common difference \( d = 4 - 9 = -5 \)[cite: 4]

Formula: \( t_n = a + (n-1)d \)[cite: 4]

Calculation:

  • \( t_{27} = 9 + (27 - 1)(-5) \)[cite: 4]
  • \( t_{27} = 9 + 26 \times (-5) \)[cite: 4]
  • \( t_{27} = 9 - 130 \)[cite: 4]
  • \( t_{27} = -121 \)[cite: 4]

Final Answer: The 27th term of the given A.P. is -121.[cite: 4]

Question 6: Find how many three-digit natural numbers are divisible by 5.

Given:

  • The three-digit natural numbers divisible by 5 are \( 100, 105, 110, \dots, 995 \).[cite: 4]
  • This sequence is an A.P.[cite: 4]
  • First term \( a = 100 \)[cite: 4]
  • Common difference \( d = 105 - 100 = 5 \)[cite: 4]
  • Last term \( t_n = 995 \)[cite: 4]

Formula: \( t_n = a + (n-1)d \)[cite: 4]

Calculation:

  • \( 995 = 100 + (n - 1)5 \)[cite: 4]
  • \( 995 - 100 = (n - 1)5 \)[cite: 4]
  • \( 895 = (n - 1)5 \)[cite: 4]
  • \( n - 1 = \frac{895}{5} \)[cite: 4]
  • \( n - 1 = 179 \)[cite: 5]
  • \( n = 179 + 1 = 180 \)[cite: 5]

Final Answer: There are 180 three-digit natural numbers which are divisible by 5.[cite: 5]

Exam Tip: Whenever asked for "how many numbers", form the A.P., identify the first and last term, and solve for \( n \) using the \( t_n \) formula.

Question 7: The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.

Given:

  • \( t_{11} = 16 \)[cite: 5]
  • \( t_{21} = 29 \)[cite: 5]

Formula: \( t_n = a + (n-1)d \)[cite: 5]

Calculation:

  • For the 11th term: \( t_{11} = a + (11 - 1)d \) → \( 16 = a + 10d \). Let this be Equation (i): \( a + 10d = 16 \).[cite: 5]
  • For the 21st term: \( t_{21} = a + (21 - 1)d \) → \( 29 = a + 20d \). Let this be Equation (ii): \( a + 20d = 29 \).[cite: 5]
  • Subtracting equation (i) from (ii):
    \( (a + 20d) - (a + 10d) = 29 - 16 \)
    \( 10d = 13 \)[cite: 6]
    \( d = \frac{13}{10} \)[cite: 6]
  • Substituting \( d = \frac{13}{10} \) in equation (i):
    \( a + 10(\frac{13}{10}) = 16 \)[cite: 6]
    \( a + 13 = 16 \)[cite: 6]
    \( a = 16 - 13 = 3 \)[cite: 6]
  • Now, to find the 41st term:
    \( t_{41} = 3 + (41 - 1)(\frac{13}{10}) \)[cite: 6]
    \( t_{41} = 3 + 40 \times \frac{13}{10} \)[cite: 6]
    \( t_{41} = 3 + 52 \)[cite: 6]
    \( t_{41} = 55 \)[cite: 6]

Final Answer: The 41st term of the A.P. is 55.[cite: 6]

Question 8: 11, 8, 5, 2, ... In this A.P. which term is number -151?

Given:

  • The A.P. is \( 11, 8, 5, 2, \dots \)[cite: 6]
  • \( a = 11 \)[cite: 6]
  • \( d = 8 - 11 = -3 \)[cite: 6]
  • Let the nth term be -151, so \( t_n = -151 \)[cite: 6]

Formula: \( t_n = a + (n-1)d \)[cite: 6]

Calculation:

  • \( -151 = 11 + (n - 1)(-3) \)[cite: 6]
  • \( -151 - 11 = (n - 1)(-3) \)[cite: 6]
  • \( -162 = (n - 1)(-3) \)[cite: 6]
  • \( n - 1 = \frac{-162}{-3} \)[cite: 6, 7]
  • \( n - 1 = 54 \)[cite: 7]
  • \( n = 54 + 1 = 55 \)[cite: 7]

Final Answer: The 55th term of the given A.P. is -151.[cite: 7]

Question 9: In the natural numbers from 10 to 250, how many are divisible by 4?

Given:

  • The natural numbers from 10 to 250 divisible by 4 form the sequence: \( 12, 16, 20, \dots, 248 \).[cite: 7]
  • This sequence is an A.P.[cite: 7]
  • \( a = 12 \)[cite: 7]
  • \( d = 16 - 12 = 4 \)[cite: 7]
  • Let the number of terms be \( n \), so \( t_n = 248 \).[cite: 7]

Formula: \( t_n = a + (n-1)d \)[cite: 7]

Calculation:

  • \( 248 = 12 + (n - 1)4 \)[cite: 7]
  • \( 248 - 12 = (n - 1)4 \)[cite: 7]
  • \( 236 = (n - 1)4 \)[cite: 7]
  • \( n - 1 = \frac{236}{4} \)[cite: 7]
  • \( n - 1 = 59 \)[cite: 7]
  • \( n = 59 + 1 = 60 \)[cite: 7]

Final Answer: There are 60 natural numbers from 10 to 250 which are divisible by 4.[cite: 7]

Question 10: In an A.P., the 17th term is 7 more than its 10th term. Find the common difference.

Given Condition: \( t_{17} = t_{10} + 7 \)[cite: 8]

Formula: \( t_n = a + (n-1)d \)[cite: 8]

Calculation:

  • \( a + (17 - 1)d = a + (10 - 1)d + 7 \)[cite: 8]
  • \( a + 16d = a + 9d + 7 \)[cite: 8]
  • \( a + 16d - a - 9d = 7 \)[cite: 8]
  • \( 7d = 7 \)[cite: 8]
  • \( d = \frac{7}{7} = 1 \)[cite: 8]

Final Answer: The common difference is 1.[cite: 8]

Application Question 1: Kabir's Height

Problem Statement: Kabir's mother keeps a record of his height on each birthday. When he was one year old, his height was 70 cm, at 2 years he was 80 cm tall and 3 years he was 90 cm tall. His aunt Meera was studying in the 10th class. She said, "it seems like Kabir's height grows in Arithmetic Progression". Assuming this, she calculated how tall Kabir will be at the age of 15 years when he is in 10th. You too assume that Kabir grows in A.P. and find out his height at the age of 15 years.[cite: 8]

Given:

  • Height at 1 year = 70 cm (\( a = 70 \))[cite: 8]
  • Height at 2 years = 80 cm[cite: 8]
  • Height at 3 years = 90 cm[cite: 8]
  • The heights form an A.P. Common difference \( d = 80 - 70 = 10 \).[cite: 8]
  • We need to find his height at age 15 (\( t_{15} \)).[cite: 8]

Formula: \( t_n = a + (n-1)d \)[cite: 8]

Calculation:

  • \( t_{15} = 70 + (15 - 1)10 \)[cite: 8]
  • \( t_{15} = 70 + 14 \times 10 = 70 + 140 \)[cite: 8]
  • \( t_{15} = 210 \)[cite: 9]

Final Answer: The height of Kabir at the age of 15 years will be 210 cm.[cite: 9]

Application Question 2: Testing Sequence Terms

Problem Statement: Is 5, 8, 11, 14, ... an A.P.? If so then what will be the 100th term? Check whether 92 is in this A.P.? Is number 61 in this A.P.?[cite: 9]

Part i: Is it an A.P.?

  • Given: \( 5, 8, 11, 14, \dots \)[cite: 9]
  • \( t_1 = 5 \), \( t_2 = 8 \), \( t_3 = 11 \), \( t_4 = 14 \)[cite: 9]
  • \( t_2 - t_1 = 8 - 5 = 3 \)[cite: 9]
  • \( t_3 - t_2 = 11 - 8 = 3 \)[cite: 9]
  • \( t_4 - t_3 = 14 - 11 = 3 \)[cite: 9]

The difference between two consecutive terms is constant (\( d = 3 \)). Therefore, the given sequence is an A.P.[cite: 9]

Part ii: What will be the 100th term?

  • \( t_{100} = 5 + (100 - 1)3 \) (using \( a = 5 \), \( d = 3 \))[cite: 9]
  • \( t_{100} = 5 + 99 \times 3 \)[cite: 9]
  • \( t_{100} = 5 + 297 \)[cite: 9]
  • \( t_{100} = 302 \)[cite: 9]

The 100th term is 302.[cite: 9]

Part iii: Check whether 92 is in this A.P.

  • Let \( t_n = 92 \)[cite: 9]
  • \( 92 = 5 + (n - 1)3 \)[cite: 9]
  • \( 92 = 5 + 3n - 3 \)[cite: 9]
  • \( 92 = 2 + 3n \)[cite: 9]
  • \( 90 = 3n \)[cite: 9]
  • \( n = \frac{90}{3} = 30 \)[cite: 9, 10]

Since \( n \) is a natural number, 92 is the 30th term of the given A.P.[cite: 10]

Part iv: Check whether 61 is in this A.P.

  • Let \( t_n = 61 \)[cite: 10]
  • \( 61 = 5 + (n - 1)3 \)[cite: 10]
  • \( 61 = 5 + 3n - 3 \)[cite: 10]
  • \( 61 = 2 + 3n \)[cite: 10]
  • \( 61 - 2 = 3n \)[cite: 10]
  • \( 59 = 3n \)[cite: 10]
  • \( n = \frac{59}{3} \)[cite: 10]

But, \( n \) must be a natural number, and 59 is not divisible by 3. Therefore, 61 is not in the given A.P.[cite: 10]

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