Maharashtra Board Class 10 Maths Chapter 3 Arithmetic Progression Set 3.1 Solutions

Question 1. Which of the following sequences are A.P.? If they are A.P. find the common difference.
(i) 2, 4, 6, 8, ...
(ii) \( 2, \frac{5}{2}, 3, \frac{7}{2}, \dots \)
(iii) -10, -6, -2, 2, ...
(iv) 0.3, 0.33, 0.333, ...
(v) 0, -4, -8, -12, ...
(vi) \( -\frac{1}{5}, -\frac{1}{5}, -\frac{1}{5}, \dots \)
(vii) \( 3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots \)
(viii) 127, 132, 137, ...
Answer:
Solution:
(i) The given sequence is 2, 4, 6, 8,...
Here, t₁ = 2, t₂ = 4, t₃ = 6, t₄ = 8
\( \therefore t_2 - t_1 = 4 - 2 = 2 \)
\( t_3 - t_2 = 6 - 4 = 2 \)
\( t_4 - t_3 = 8 - 6 = 2 \)
\( \therefore t_2 - t_1 = t_3 - t_2 = \dots = 2 = d = \text{constant} \)
The difference between two consecutive terms is constant.
\( \therefore \) The given sequence is an A.P. and common difference (d) = 2.
(ii) The given sequence is \( 2, \frac{5}{2}, 3, \frac{7}{2}, \dots \)
Here, \( t_1 = 2, t_2 = \frac{5}{2}, t_3 = 3, t_4 = \frac{7}{2} \)
\( \therefore t_2 - t_1 = \frac{5}{2} - 2 = \frac{5-4}{2} = \frac{1}{2} \)
\( t_3 - t_2 = 3 - \frac{5}{2} = \frac{6-5}{2} = \frac{1}{2} \)
\( t_4 - t_3 = \frac{7}{2} - 3 = \frac{7-6}{2} = \frac{1}{2} \)
\( \therefore t_2 - t_1 = t_3 - t_2 = \dots = \frac{1}{2} = d = \text{constant} \)
The difference between two consecutive terms is constant.
\( \therefore \) The given sequence is an A.P. and common difference \( (d) = \frac{1}{2} \).
(iii) The given sequence is -10, -6, -2, 2,...
Here, t₁ = -10, t₂ = -6, t₃ = -2, t₄ = 2
\( \therefore t_2 - t_1 = -6 - (-10) = -6 + 10 = 4 \)
\( t_3 - t_2 = -2 - (-6) = -2 + 6 = 4 \)
\( t_4 - t_3 = 2 - (-2) = 2 + 2 = 4 \)
\( \therefore t_2 - t_1 = t_3 - t_2 = \dots = 4 = d = \text{constant} \)
The difference between two consecutive terms is constant.
\( \therefore \) The given sequence is an A.P. and common difference (d) = 4.
(iv) The given sequence is 0.3, 0.33, 0.333,...
Here, t₁ = 0.3, t₂ = 0.33, t₃ = 0.333
\( \therefore t_2 - t_1 = 0.33 - 0.3 = 0.03 \)
\( t_3 - t_2 = 0.333 - 0.33 = 0.003 \)
\( \therefore t_2 - t_1 \ne t_3 - t_2 \)
The difference between two consecutive terms is not constant.
\( \therefore \) The given sequence is not an A.P.
(v) The given sequence is 0, -4, -8, -12,...
Here, t₁ = 0, t₂ = -4, t₃ = -8, t₄ = -12
\( \therefore t_2 - t_1 = -4 - 0 = -4 \)
\( t_3 - t_2 = -8 - (-4) = -8 + 4 = -4 \)
\( t_4 - t_3 = -12 - (-8) = -12 + 8 = -4 \)
\( \therefore t_2 - t_1 = t_3 - t_2 = \dots = -4 = d = \text{constant} \)
The difference between two consecutive terms is constant.
\( \therefore \) The given sequence is an A.P. and common difference (d) = -4.
(vi) The given sequence is \( -\frac{1}{5}, -\frac{1}{5}, -\frac{1}{5}, \dots \)
Here, \( t_1 = -\frac{1}{5}, t_2 = -\frac{1}{5}, t_3 = -\frac{1}{5} \)
\( \therefore t_2 - t_1 = -\frac{1}{5} - (-\frac{1}{5}) = -\frac{1}{5} + \frac{1}{5} = 0 \)
\( t_3 - t_2 = -\frac{1}{5} - (-\frac{1}{5}) = -\frac{1}{5} + \frac{1}{5} = 0 \)
\( \therefore t_2 - t_1 = t_3 - t_2 = \dots = 0 = d = \text{constant} \)
The difference between two consecutive terms is constant.
\( \therefore \) The given sequence is an A.P. and common difference (d) = 0.
(vii) The given sequence is \( 3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots \)
Here, \( t_1 = 3, t_2 = 3+\sqrt{2}, t_3 = 3+2\sqrt{2}, t_4 = 3+3\sqrt{2} \)
\( \therefore t_2 - t_1 = (3+\sqrt{2}) - 3 = \sqrt{2} \)
\( t_3 - t_2 = (3+2\sqrt{2}) - (3+\sqrt{2}) = \sqrt{2} \)
\( t_4 - t_3 = (3+3\sqrt{2}) - (3+2\sqrt{2}) = \sqrt{2} \)
\( \therefore t_2 - t_1 = t_3 - t_2 = \dots = \sqrt{2} = d = \text{constant} \)
The difference between two consecutive terms is constant.
\( \therefore \) The given sequence is an A.P. and common difference \( (d) = \sqrt{2} \).
(viii) The given sequence is 127, 132, 137,...
Here, t₁ = 127, t₂ = 132, t₃ = 137
\( \therefore t_2 - t_1 = 132 - 127 = 5 \)
\( t_3 - t_2 = 137 - 132 = 5 \)
\( \therefore t_2 - t_1 = t_3 - t_2 = \dots = 5 = d = \text{constant} \)
The difference between two consecutive terms is constant.
\( \therefore \) The given sequence is an A.P. and common difference (d) = 5.
In simple words: To check if a sequence is an Arithmetic Progression (A.P.), calculate the difference between consecutive terms. If this difference (common difference 'd') is constant throughout the sequence, then it is an A.P.

🎯 Exam Tip: Clearly show the calculation of differences between consecutive terms to demonstrate constancy (or lack thereof) for full marks. Identify 'd' correctly if it's an A.P.

Question 2. Write an A.P. whose first term is a and common difference is d in each of the following.
(i) a = 10, d = 5
(ii) a = -3, d = 0
(iii) \( a = -7, d = \frac{1}{2} \)
(iv) a = -1.25, d = 3
(v) a = 6, d = -3
(vi) a = -19, d = -4
Answer:
Solution:
(i) a = 10, d = 5
\( \therefore t_1 = a = 10 \)
\( t_2 = t_1 + d = 10 + 5 = 15 \)
\( t_3 = t_2 + d = 15 + 5 = 20 \)
\( t_4 = t_3 + d = 20 + 5 = 25 \)
\( \therefore \) The required A.P. is 10, 15, 20, 25,...
(ii) a = -3, d = 0
\( \therefore t_1 = a = -3 \)
\( t_2 = t_1 + d = -3 + 0 = -3 \)
\( t_3 = t_2 + d = -3 + 0 = -3 \)
\( t_4 = t_3 + d = -3 + 0 = -3 \)
\( \therefore \) The required A.P. is -3, -3, -3, -3,...
(iii) \( a = -7, d = -\frac{1}{2} \)
\( \therefore t_1 = a = -7 \)
\( t_2 = t_1 + d = -7 + \left(-\frac{1}{2}\right) = -7 - \frac{1}{2} = \frac{-14-1}{2} = -\frac{15}{2} = -7.5 \)
\( t_3 = t_2 + d = -7.5 + \left(-\frac{1}{2}\right) = -7.5 - 0.5 = -8 \)
\( t_4 = t_3 + d = -8 + \left(-\frac{1}{2}\right) = -8 - 0.5 = -8.5 \)
\( \therefore \) The required A.P. is -7, -7.5, -8, -8.5,...
(iv) a = -1.25, d = 3
\( t_1 = a = -1.25 \)
\( t_2 = t_1 + d = -1.25 + 3 = 1.75 \)
\( t_3 = t_2 + d = 1.75 + 3 = 4.75 \)
\( t_4 = t_3 + d = 4.75 + 3 = 7.75 \)
\( \therefore \) The required A.P. is -1.25, 1.75, 4.75, 7.75,...
(v) a = 6, d = -3
\( \therefore t_1 = a = 6 \)
\( t_2 = t_1 + d = 6 - 3 = 3 \)
\( t_3 = t_2 + d = 3 - 3 = 0 \)
\( t_4 = t_3 + d = 0 - 3 = -3 \)
\( \therefore \) The required A.P. is 6, 3, 0, -3,...
(vi) a = -19, d = -4
\( t_1 = a = -19 \)
\( t_2 = t_1 + d = -19 - 4 = -23 \)
\( t_3 = t_2 + d = -23 - 4 = -27 \)
\( t_4 = t_3 + d = -27 - 4 = -31 \)
\( \therefore \) The required A.P. is -19, -23, -27, -31,...
In simple words: An Arithmetic Progression is formed by adding the common difference 'd' to each preceding term, starting from the first term 'a'. Each term is \(t_n = a + (n-1)d\).

🎯 Exam Tip: Ensure accurate calculations for positive and negative numbers, especially when 'd' is negative or a fraction. Clearly state the first few terms of the generated A.P.

Question 3. Find the first term and common difference for each of the A.P.
(i) 5, 1, -3, -7, ...
(ii) 0.6, 0.9, 1.2, 1.5, ...
(iii) 127, 135, 143, 151, ...
(iv) \( \frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \dots \)
Answer:
Solution:
(i) The given A.P. is 5, 1,-3,-7,...
Here, t₁ = 5, t₂ = 1
\( \therefore a = t_1 = 5 \) and
\( d = t_2 - t_1 = 1 - 5 = -4 \)
\( \therefore \) first term (a) = 5,
common difference (d) = -4
(ii) The given A.P. is 0.6, 0.9, 1.2, 1.5,...
Here, t₁ = 0.6, t₂ = 0.9
\( \therefore a = t_1 = 0.6 \) and
\( d = t_2 - t_1 = 0.9 - 0.6 = 0.3 \)
\( \therefore \) first term (a) = 0.6,
common difference (d) = 0.3
(iii) The given A.P. is 127, 135, 143, 151,...
Here, t₁ = 127, t₂ = 135
\( \therefore a = t_1 = 127 \) and
\( d = t_2 - t_1 = 135 - 127 = 8 \)
\( \therefore \) first term (a) = 127,
common difference (d) = 8
(iv) The given A.P. is \( \frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \dots \)
Here, \( t_1 = \frac{1}{4}, t_2 = \frac{3}{4} \)
\( \therefore a = t_1 = \frac{1}{4} \) and
\( d = t_2 - t_1 = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \)
\( \therefore \) first term (a) = \( \frac{1}{4} \),
common difference (d) = \( \frac{1}{2} \)
In simple words: The first term 'a' is simply the first number in the sequence. The common difference 'd' is found by subtracting any term from its succeeding term.

🎯 Exam Tip: Always show the calculation for 'd' by subtracting \(t_1\) from \(t_2\). Be careful with fractions and decimals in your calculations.

Question 1. Complete the given pattern. Look at the pattern of the numbers. Try to find a rule to obtain the next number from its preceding number. Write the next numbers. (Textbook pg, no. 55 and 56)
Answer:
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पैटर्न वृत्त (circles) की संख्या को दर्शाता है। प्रत्येक अगले पैटर्न में क्षैतिज और ऊर्ध्वाधर पंक्तियों में एक वृत्त जोड़ा जाता है, जिससे कुल वृत्तों की संख्या बढ़ती जाती है।

Every pattern is formed by adding a circle in horizontal and vertical rows to the preceding pattern.
\( \therefore \) The sequence for the above pattern is 1, 3, 5, 7, 9, 11, 13, 15, 17,....
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पैटर्न त्रिभुज (triangles) की संख्या को दर्शाता है। प्रत्येक अगले पैटर्न में क्षैतिज रूप से 2 त्रिभुज और ऊर्ध्वाधर रूप से 1 त्रिभुज जोड़ा जाता है, जिससे कुल त्रिभुजों की संख्या बढ़ती जाती है।

Every pattern is formed by adding 2 triangles horizontally and 1 triangle vertically to the preceding pattern.
\( \therefore \) The sequence for the above pattern is 5, 8, 11, 14, 17, 20, 23,...
In simple words: Patterns often follow a numerical rule. By observing how each term changes to the next, we can deduce the rule, which might involve adding a constant number, like in an A.P.

🎯 Exam Tip: For pattern-based questions, carefully count the elements in each step and look for a consistent arithmetic operation (addition/subtraction) or multiplication to find the rule.

Question 2. Some sequences are given below. Show the positions of the terms by t₁, t₂, t₃,...
Answer:
(i) 7, 7, 7, 7, ... Here t₁ = 7, t₂ = 7, t₃ = 7
(ii) -2, -6, -10, -14, ... Here t₁ = -2, t₂ = -6, t₃ = -10
In simple words: Each number in a sequence is called a term, and its position is denoted by \(t_n\), where 'n' is the term number (e.g., \(t_1\) for the first term, \(t_2\) for the second, and so on).

🎯 Exam Tip: Correctly identifying \(t_1, t_2, t_3\), etc., is crucial for setting up calculations in A.P. problems.

Question 3. Some sequences are given below. Check whether there is any rule among the terms. Find the similarity between two sequences. To check the rule for the terms of the sequence look at the arrangements and fill the empty boxes suitably. (Textbook pg. no. 56 and 57)
(i) 1, 4, 7, 10, 13,...
(ii) 6, 12, 18, 24,...
(iii) 3, 3, 3, 3,...
(iv) 4, 16, 64,...
(v) -1, -1.5, -2, -2.5,...
(vi) \( 1^3, 2^3, 3^3, 4^3 \)
Answer:
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पैटर्न संख्याओं के बीच '3' जोड़ने का नियम दर्शाता है। प्रत्येक संख्या में 3 जोड़ने पर अगली संख्या प्राप्त होती है, जैसे 1+3=4, 4+3=7, इत्यादि।
\[ \begin{array}{ccccccc} 1 & \xrightarrow{(+3)} & 4 & \xrightarrow{(+3)} & 7 & \xrightarrow{(+3)} & 10 \\ & & 4 & \xrightarrow{(+3)} & 7 & \xrightarrow{(+3)} & 10 \\ & & & & & & \\ 10 & \xrightarrow{(+3)} & 13 & \xrightarrow{(+3)} & 16 & \dots & \\ & & 13 & \xrightarrow{(+3)} & 16 & \dots & \end{array} \]
By adding 3 in each term, we get the next term of the sequence.
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पैटर्न संख्याओं के बीच '6' जोड़ने का नियम दर्शाता है। प्रत्येक संख्या में 6 जोड़ने पर अगली संख्या प्राप्त होती है, जैसे 6+6=12, 12+6=18, इत्यादि।
\[ \begin{array}{ccccccc} 6 & \xrightarrow{(+6)} & 12 & \xrightarrow{(+6)} & 18 & \xrightarrow{(+6)} & 24 \\ & & 12 & \xrightarrow{(+6)} & 18 & \xrightarrow{(+6)} & 24 \\ & & & & & & \\ 24 & \xrightarrow{(+6)} & 30 & \xrightarrow{(+6)} & 36 & \dots & \\ & & 30 & \xrightarrow{(+6)} & 36 & \dots & \end{array} \]
By adding 6 in each term, we get the next term of the sequence.
(iii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पैटर्न संख्याओं के बीच '0' जोड़ने का नियम दर्शाता है। प्रत्येक संख्या में 0 जोड़ने पर अगली संख्या प्राप्त होती है, जिसका अर्थ है कि संख्या स्थिर रहती है, जैसे 3+0=3।
\[ \begin{array}{ccccccc} 3 & \xrightarrow{(+0)} & 3 & \xrightarrow{(+0)} & 3 & \xrightarrow{(+0)} & 3 \\ & & 3 & \xrightarrow{(+0)} & 3 & \xrightarrow{(+0)} & 3 \\ & & & & & & \\ 3 & \xrightarrow{(+0)} & 3 & \xrightarrow{(+0)} & 3 & \dots & \\ & & 3 & \xrightarrow{(+0)} & 3 & \dots & \end{array} \]
By adding 0 in each term, we get the next term of the sequence.
(iv)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पैटर्न संख्याओं के बीच '4' से गुणा करने का नियम दर्शाता है। प्रत्येक संख्या को 4 से गुणा करने पर अगली संख्या प्राप्त होती है, जैसे 4x4=16, 16x4=64, इत्यादि।
\[ \begin{array}{ccccccc} 4 & \xrightarrow{\times 4} & 16 & \xrightarrow{\times 4} & 64 & \xrightarrow{\times 4} & 256 \\ & & 16 & \xrightarrow{\times 4} & 64 & \xrightarrow{\times 4} & 256 \\ & & & & & & \\ 256 & \xrightarrow{\times 4} & 1024 & \xrightarrow{\times 4} & 4096 & \dots & \\ & & 1024 & \xrightarrow{\times 4} & 4096 & \dots & \end{array} \]
By multiplying each term by 4, we get the next term of the sequence.
(v)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह पैटर्न संख्याओं के बीच '-0.5' जोड़ने का नियम दर्शाता है। प्रत्येक संख्या में -0.5 जोड़ने पर अगली संख्या प्राप्त होती है, जैसे (-1)+(-0.5)=-1.5, (-1.5)+(-0.5)=-2, इत्यादि।
\[ \begin{array}{ccccccc} -1 & \xrightarrow{+(-0.5)} & -1.5 & \xrightarrow{+(-0.5)} & -2 & \xrightarrow{+(-0.5)} & -2.5 \\ & & -1.5 & \xrightarrow{+(-0.5)} & -2 & \xrightarrow{+(-0.5)} & -2.5 \\ & & & & & & \\ -2.5 & \xrightarrow{+(-0.5)} & -3 & \dots & & & \\ & & -3 & \dots & & & \end{array} \]
By adding -0.5 in each term, we get the next term of the sequence.
(vi) \( 1^3, 2^3, 3^3, \dots \)
(next term)³ = (previous term +1)³
The similarity in the sequences i., ii., iii. and v. is that the next term is obtained by adding a particular fixed number to the previous term.
Note: A Geometric Progression is a sequence in which the ratio of any two consecutive terms is a constant,
i.e. in a G.P., \( \frac{t_2}{t_1} = \frac{t_3}{t_2} = \dots = \frac{t_n}{t_{n-1}} = \text{constant} \)
Sequence iv. is a geometric progression.
In simple words: Sequences can follow different rules: some add a constant number (Arithmetic Progression), some multiply by a constant number (Geometric Progression), and some follow other mathematical operations like cubing consecutive natural numbers.

🎯 Exam Tip: Differentiate between Arithmetic Progression (constant difference) and Geometric Progression (constant ratio). For other patterns, describe the rule clearly, such as operations on natural numbers.

Question 4. Write one example of finite and infinite A.P. each. (Textbook pg. no. 59)
Answer:
Finite A.P.:
Even natural numbers from 4 to 50:
4, 6, 8, ................, 50.
Infinite A. P.:
Positive multiples of 5:
5, 10, 15, ..............
In simple words: A finite A.P. has a limited number of terms, with a clear beginning and end, while an infinite A.P. continues indefinitely without an end term.

🎯 Exam Tip: To define a finite A.P., specify its start and end terms. For an infinite A.P., indicate the starting term and the common difference, followed by ellipses (...).