Class 10 Maths Part 1 Practice Set 3.1 Solutions | Maharashtra Board
Welcome to our comprehensive guide for Class 10 Maths Part 1 Practice Set 3.1. If you are a Maharashtra Board student studying Chapter 3 Arithmetic Progression, these step-by-step Textbook Exercise Solutions will help you master the core concepts. From identifying whether a sequence is an Arithmetic Progression (A.P.) to calculating the first term and common difference, our detailed breakdown is designed to help you secure full marks in your board exams. Let's dive into the solutions!
Exam Tip: To prove a sequence is an A.P., always show that the difference between at least three consecutive pairs of terms, i.e., \( t_2 - t_1 \) and \( t_3 - t_2 \), is constant. Always enclose negative numbers in brackets during subtraction to avoid careless sign errors!
Practice Set 3.1: Question 1
Which of the following sequences are A.P.? If they are A.P. find the common difference.
i. \( 2, 4, 6, 8, \dots \)
- Given Sequence: \( 2, 4, 6, 8, \dots \)
- Here: \( t_1 = 2 \), \( t_2 = 4 \), \( t_3 = 6 \), \( t_4 = 8 \)
- Calculation:
\( t_2 - t_1 = 4 - 2 = 2 \)
\( t_3 - t_2 = 6 - 4 = 2 \)
\( t_4 - t_3 = 8 - 6 = 2 \) - Since \( t_2 - t_1 = t_3 - t_2 = \dots = 2 = d \) (constant).
- The difference between two consecutive terms is constant.
- Final Answer: The given sequence is an A.P. and the common difference \( (d) = 2 \).
ii. \( 2, \frac{5}{2}, 3, \frac{7}{2}, \dots \)
- Given Sequence: \( 2, \frac{5}{2}, 3, \frac{7}{2}, \dots \)
- Here: \( t_1 = 2 \), \( t_2 = \frac{5}{2} \), \( t_3 = 3 \), \( t_4 = \frac{7}{2} \)
- Calculation:
\( t_2 - t_1 = \frac{5}{2} - 2 = \frac{5 - 4}{2} = \frac{1}{2} \)
\( t_3 - t_2 = 3 - \frac{5}{2} = \frac{6 - 5}{2} = \frac{1}{2} \)
\( t_4 - t_3 = \frac{7}{2} - 3 = \frac{7 - 6}{2} = \frac{1}{2} \) - Since \( t_2 - t_1 = t_3 - t_2 = \dots = \frac{1}{2} = d \) (constant).
- The difference between two consecutive terms is constant.
- Final Answer: The given sequence is an A.P. and the common difference \( (d) = \frac{1}{2} \).
iii. \( -10, -6, -2, 2, \dots \)
- Given Sequence: \( -10, -6, -2, 2, \dots \)
- Here: \( t_1 = -10 \), \( t_2 = -6 \), \( t_3 = -2 \), \( t_4 = 2 \)
- Calculation:
\( t_2 - t_1 = -6 - (-10) = -6 + 10 = 4 \)
\( t_3 - t_2 = -2 - (-6) = -2 + 6 = 4 \)
\( t_4 - t_3 = 2 - (-2) = 2 + 2 = 4 \) - Since \( t_2 - t_1 = t_3 - t_2 = \dots = 4 = d \) (constant).
- The difference between two consecutive terms is constant.
- Final Answer: The given sequence is an A.P. and the common difference \( (d) = 4 \).
iv. \( 0.3, 0.33, 0.333, \dots \)
- Given Sequence: \( 0.3, 0.33, 0.333, \dots \)
- Here: \( t_1 = 0.3 \), \( t_2 = 0.33 \), \( t_3 = 0.333 \)
- Calculation:
\( t_2 - t_1 = 0.33 - 0.3 = 0.03 \)
\( t_3 - t_2 = 0.333 - 0.33 = 0.003 \) - Since \( t_2 - t_1 \neq t_3 - t_2 \).
- The difference between two consecutive terms is not constant.
- Final Answer: The given sequence is not an A.P.
v. \( 0, -4, -8, -12, \dots \)
- Given Sequence: \( 0, -4, -8, -12, \dots \)
- Here: \( t_1 = 0 \), \( t_2 = -4 \), \( t_3 = -8 \), \( t_4 = -12 \)
- Calculation:
\( t_2 - t_1 = -4 - 0 = -4 \)
\( t_3 - t_2 = -8 - (-4) = -8 + 4 = -4 \)
\( t_4 - t_3 = -12 - (-8) = -12 + 8 = -4 \) - Since \( t_2 - t_1 = t_3 - t_2 = \dots = -4 = d \) (constant).
- The difference between two consecutive terms is constant.
- Final Answer: The given sequence is an A.P. and the common difference \( (d) = -4 \).
vi. \( -\frac{1}{5}, -\frac{1}{5}, -\frac{1}{5}, \dots \)
- Given Sequence: \( -\frac{1}{5}, -\frac{1}{5}, -\frac{1}{5}, \dots \)
- Here: \( t_1 = -\frac{1}{5} \), \( t_2 = -\frac{1}{5} \), \( t_3 = -\frac{1}{5} \)
- Calculation:
\( t_2 - t_1 = -\frac{1}{5} - \left(-\frac{1}{5}\right) = -\frac{1}{5} + \frac{1}{5} = 0 \)
\( t_3 - t_2 = -\frac{1}{5} - \left(-\frac{1}{5}\right) = -\frac{1}{5} + \frac{1}{5} = 0 \) - Since \( t_2 - t_1 = t_3 - t_2 = \dots = 0 = d \) (constant).
- The difference between two consecutive terms is constant.
- Final Answer: The given sequence is an A.P. and the common difference \( (d) = 0 \).
vii. \( 3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots \)
- Given Sequence: \( 3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots \)
- Here: \( t_1 = 3 \), \( t_2 = 3+\sqrt{2} \), \( t_3 = 3+2\sqrt{2} \), \( t_4 = 3+3\sqrt{2} \)
- Calculation:
\( t_2 - t_1 = (3+\sqrt{2}) - 3 = \sqrt{2} \)
\( t_3 - t_2 = (3+2\sqrt{2}) - (3+\sqrt{2}) = \sqrt{2} \)
\( t_4 - t_3 = (3+3\sqrt{2}) - (3+2\sqrt{2}) = \sqrt{2} \) - Since \( t_2 - t_1 = t_3 - t_2 = \dots = \sqrt{2} = d \) (constant).
- The difference between two consecutive terms is constant.
- Final Answer: The given sequence is an A.P. and the common difference \( (d) = \sqrt{2} \).
viii. \( 127, 132, 137, \dots \)
- Given Sequence: \( 127, 132, 137, \dots \)
- Here: \( t_1 = 127 \), \( t_2 = 132 \), \( t_3 = 137 \)
- Calculation:
\( t_2 - t_1 = 132 - 127 = 5 \)
\( t_3 - t_2 = 137 - 132 = 5 \) - Since \( t_2 - t_1 = t_3 - t_2 = \dots = 5 = d \) (constant).
- The difference between two consecutive terms is constant.
- Final Answer: The given sequence is an A.P. and the common difference \( (d) = 5 \).
Practice Set 3.1: Question 2
Write an A.P. whose first term is \( a \) and common difference is \( d \) in each of the following.
Key Formula: The terms of an A.P. can be found using the sequence: \( t_1 = a \), \( t_2 = t_1 + d \), \( t_3 = t_2 + d \), \( \dots \)
i. \( a = 10 \), \( d = 5 \)
- Given: \( a = 10 \), \( d = 5 \)
- Calculation:
\( t_1 = a = 10 \)
\( t_2 = t_1 + d = 10 + 5 = 15 \)
\( t_3 = t_2 + d = 15 + 5 = 20 \)
\( t_4 = t_3 + d = 20 + 5 = 25 \) - Final Answer: The required A.P. is 10, 15, 20, 25, ...
ii. \( a = -3 \), \( d = 0 \)
- Given: \( a = -3 \), \( d = 0 \)
- Calculation:
\( t_1 = a = -3 \)
\( t_2 = t_1 + d = -3 + 0 = -3 \)
\( t_3 = t_2 + d = -3 + 0 = -3 \)
\( t_4 = t_3 + d = -3 + 0 = -3 \) - Final Answer: The required A.P. is -3, -3, -3, -3, ...
iii. \( a = -7 \), \( d = \frac{1}{2} \)
- Given: \( a = -7 \), \( d = \frac{1}{2} \) (or \( 0.5 \))
- Calculation:
\( t_1 = a = -7 \)
\( t_2 = t_1 + d = -7 + \frac{1}{2} = \frac{-14 + 1}{2} = \frac{-13}{2} = -6.5 \)
\( t_3 = t_2 + d = -6.5 + 0.5 = -6 \)
\( t_4 = t_3 + d = -6 + \frac{1}{2} = \frac{-12 + 1}{2} = \frac{-11}{2} = -5.5 \) - Final Answer: The required A.P. is -7, -6.5, -6, -5.5, ...
iv. \( a = -1.25 \), \( d = 3 \)
- Given: \( a = -1.25 \), \( d = 3 \)
- Calculation:
\( t_1 = a = -1.25 \)
\( t_2 = t_1 + d = -1.25 + 3 = 1.75 \)
\( t_3 = t_2 + d = 1.75 + 3 = 4.75 \)
\( t_4 = t_3 + d = 4.75 + 3 = 7.75 \) - Final Answer: The required A.P. is -1.25, 1.75, 4.75, 7.75, ...
v. \( a = 6 \), \( d = -3 \)
- Given: \( a = 6 \), \( d = -3 \)
- Calculation:
\( t_1 = a = 6 \)
\( t_2 = t_1 + d = 6 + (-3) = 3 \)
\( t_3 = t_2 + d = 3 + (-3) = 0 \)
\( t_4 = t_3 + d = 0 + (-3) = -3 \) - Final Answer: The required A.P. is 6, 3, 0, -3, ...
vi. \( a = -19 \), \( d = -4 \)
- Given: \( a = -19 \), \( d = -4 \)
- Calculation:
\( t_1 = a = -19 \)
\( t_2 = t_1 + d = -19 + (-4) = -23 \)
\( t_3 = t_2 + d = -23 + (-4) = -27 \)
\( t_4 = t_3 + d = -27 + (-4) = -31 \) - Final Answer: The required A.P. is -19, -23, -27, -31, ...
Practice Set 3.1: Question 3
Find the first term and common difference for each of the A.P.
i. \( 5, 1, -3, -7, \dots \)
- Given A.P.: \( 5, 1, -3, -7, \dots \)
- Here: \( t_1 = 5 \), \( t_2 = 1 \)
- Calculation:
First term \( a = t_1 = 5 \)
Common difference \( d = t_2 - t_1 = 1 - 5 = -4 \) - Final Answer: First term \( (a) = 5 \), common difference \( (d) = -4 \).
ii. \( 0.6, 0.9, 1.2, 1.5, \dots \)
- Given A.P.: \( 0.6, 0.9, 1.2, 1.5, \dots \)
- Here: \( t_1 = 0.6 \), \( t_2 = 0.9 \)
- Calculation:
First term \( a = t_1 = 0.6 \)
Common difference \( d = t_2 - t_1 = 0.9 - 0.6 = 0.3 \) - Final Answer: First term \( (a) = 0.6 \), common difference \( (d) = 0.3 \).
iii. \( 127, 135, 143, 151, \dots \)
- Given A.P.: \( 127, 135, 143, 151, \dots \)
- Here: \( t_1 = 127 \), \( t_2 = 135 \)
- Calculation:
First term \( a = t_1 = 127 \)
Common difference \( d = t_2 - t_1 = 135 - 127 = 8 \) - Final Answer: First term \( (a) = 127 \), common difference \( (d) = 8 \).
iv. \( \frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \dots \)
- Given A.P.: \( \frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \dots \)
- Here: \( t_1 = \frac{1}{4} \), \( t_2 = \frac{3}{4} \)
- Calculation:
First term \( a = t_1 = \frac{1}{4} \)
Common difference \( d = t_2 - t_1 = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \) - Final Answer: First term \( (a) = \frac{1}{4} \), common difference \( (d) = \frac{1}{2} \).
Additional Textbook Questions (Page 55 & 56)
Question 1: Complete the given pattern
i. Circle Pattern
| Pattern Number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|
| Number of circles | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 | 17 |
Rule: Every pattern is formed by adding a circle in horizontal and vertical rows to the preceding pattern. The sequence is 1, 3, 5, 7, 9, 11, 13, 15, 17, ...
ii. Triangle Pattern
| Pattern Number | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Number of triangles | 5 | 8 | 11 | 14 | 17 | 20 | 23 |
Rule: Every pattern is formed by adding 2 triangles horizontally and 1 triangle vertically to the preceding pattern. The sequence is 5, 8, 11, 14, 17, 20, 23, ...
Question 2: Show the positions of the terms by \( t_1, t_2, t_3, \dots \)
- i. \( 7, 7, 7, 7, \dots \)
Here \( t_1 = 7 \), \( t_2 = 7 \), \( t_3 = 7 \), \( \dots \) - ii. \( -2, -6, -10, -14, \dots \)
Here \( t_1 = -2 \), \( t_2 = -6 \), \( t_3 = -10 \), \( \dots \)
Question 3: Check whether there is any rule among the terms
- i. \( 1, 4, 7, 10, 13, \dots \)
Rule: By adding 3 to each term, we get the next term. - ii. \( 6, 12, 18, 24, \dots \)
Rule: By adding 6 to each term, we get the next term. - iii. \( 3, 3, 3, 3, \dots \)
Rule: By adding 0 to each term, we get the next term. - iv. \( 4, 16, 64, \dots \)
Rule: By multiplying each term by 4, we get the next term. (Note: This is a Geometric Progression). - v. \( -1, -1.5, -2, -2.5, \dots \)
Rule: By adding -0.5 to each term, we get the next term. - vi. \( 1^3, 2^3, 3^3, 4^3, \dots \)
Rule: (Next term) = \( (\text{previous term's base} + 1)^3 \).
Conclusion: The similarity in sequences i, ii, iii, and v is that the next term is obtained by adding a particular fixed number to the previous term (Arithmetic Progression).
Question 4: Write one example of finite and infinite A.P. each.
- Finite A.P.: Even natural numbers from 4 to 50:
4, 6, 8, \dots, 50 - Infinite A.P.: Positive multiples of 5:
5, 10, 15, \dots