Maharashtra Board Class 10 Maths Chapter 3 Arithmetic Progression Set 3.2 Solutions

Question 1. Write the correct number in the given boxes from the following A.P.
(i) 1, 8, 15, 22, ...
Here a = , t₁ = , t₂ = , t₃ = , ...
t₂-t₁=
t₃-t₂=-= ∴d=
(ii) 3, 6, 9, 12, ...
Here t₁ = , t₂ = , t₃ = , t₄ = , ...
t₂-t₁= , t₃-t₂= ∴d=
(iii) -3, -8, -13, -18, ...
Here t₁ = , t₂ = , t₃ = , t₄ = , ...
t₂-t₁= , t₃-t₂= ∴a= , d=
(iv) 70, 60, 50, 40, ...
Here t₁ = , t₂ = , t₃ = , ...
∴a= , d=

Answer:
(i) 1, 8, 15, 22, ...
Here a = \( \boxed{1} \), t₁ = \( \boxed{1} \), t₂ = \( \boxed{8} \), t₃ = \( \boxed{15} \), ...
t₂ - t₁ = \( \boxed{8} - \boxed{1} = \boxed{7} \)
t₃ - t₂ = \( \boxed{15} - \boxed{8} = \boxed{7} \)
\( \implies \) d = \( \boxed{7} \)
(ii) 3, 6, 9, 12, ...
Here t₁ = \( \boxed{3} \), t₂ = \( \boxed{6} \), t₃ = \( \boxed{9} \), t₄ = \( \boxed{12} \), ...
t₂ - t₁ = \( \boxed{3} \), t₃ - t₂ = \( \boxed{3} \)
\( \implies \) d = \( \boxed{3} \)
(iii) -3, -8, -13, -18, ...
Here t₁ = \( \boxed{-3} \), t₂ = \( \boxed{-8} \), t₃ = \( \boxed{-13} \), t₄ = \( \boxed{-18} \), ...
t₂ - t₁ = \( \boxed{-5} \), t₃ - t₂ = \( \boxed{-5} \)
\( \implies \) a = \( \boxed{-3} \), d = \( \boxed{-5} \)
(iv) 70, 60, 50, 40, ...
Here t₁ = \( \boxed{70} \), t₂ = \( \boxed{60} \), t₃ = \( \boxed{50} \), ...
\( \implies \) a = \( \boxed{70} \), d = \( \boxed{-10} \)
In simple words: For an Arithmetic Progression (A.P.), the first term is denoted by 'a' (or t₁), and the common difference 'd' is found by subtracting any term from its preceding term (e.g., t₂ - t₁ or t₃ - t₂). If the difference is constant, it's an A.P.

🎯 Exam Tip: Always verify the common difference 'd' by checking at least two pairs of consecutive terms to ensure the given sequence is indeed an A.P. Correctly identifying 'a' and 'd' is fundamental for all subsequent A.P. calculations.

Question 2. Decide whether following sequence is an A.P., if so find the 20th term of the progression. -12, -5, 2, 9,16, 23,30,...
Answer:
Solution:
i. The given sequence is
-12, -5, 2, 9, 16, 23,30,...
Here, t₁ = -12, t₂ = -5, t₃ = 2, t₄ = 9
\( \implies \) t₂ - t₁ = -5 - (-12) = -5 + 12 = 7
t₃ - t₂ = 2 - (-5) = 2 + 5 = 7
\( \implies \) t₄ - t₃ = 9 - 2 = 7
\( \implies \) t₂ - t₁ = t₃ - t₂ = ... = 7 = d = constant
The difference between two consecutive terms is constant.
\( \implies \) The given sequence is an A.P.
ii. t_n = a + (n - 1)d
\( \implies \) t₂₀ = -12 + (20 - 1)7 ...[∵a = -12, d = 7]
= -12 + 19 × 7
= -12 + 133
\( \implies \) t₂₀ = 121
\( \implies \) 20th term of the given A.P. is 121.
In simple words: First, calculate the difference between consecutive terms to confirm if it's a constant, hence an A.P. If it is, use the formula t_n = a + (n - 1)d to find any specific term, substituting the first term 'a', common difference 'd', and the term number 'n'.

🎯 Exam Tip: When determining if a sequence is an A.P., always calculate and compare the differences between at least two pairs of consecutive terms. For finding the nth term, clearly state the values of 'a', 'd', and 'n' before applying the formula t_n = a + (n - 1)d.

Question 3. Given Arithmetic Progression is 12, 16, 20, 24, ... Find the 24th term of this progression.
Answer:
Solution:
The given A.P. is 12, 16, 20, 24,...
Here, a = 12, d = 16 - 12 = 4 Since,
t_n = a + (n-1)d
\( \implies \) t₂₄ = 12 + (24 - 1)4
= 12 + 23 × 4
= 12 + 92
\( \implies \) t₂₄ = 104
\( \implies \) 24th term of the given A.P. is 104.
In simple words: To find a specific term in an A.P., identify the first term 'a' and the common difference 'd'. Then, use the formula t_n = a + (n-1)d, where 'n' is the position of the term you want to find.

🎯 Exam Tip: Clearly write down the values of 'a' and 'd' at the beginning of your solution. Show each step of the calculation when applying the t_n formula to avoid errors and secure full marks.

Question 4. Find the 19th term of the following A.P. 7,13,19,25.....
Answer:
Solution:
The given A.P. is 7, 13, 19, 25,...
Here, a = 7, d = 13 - 7 = 6
Since, t_n = a + (n - 1)d
\( \implies \) t₁₉ = 7 + (19 - 1)6
= 7 + 18 × 6
= 7 + 108
\( \implies \) t₁₉ = 115
\( \implies \) 19th term of the given A.P. is 115.
In simple words: To find the 19th term, first identify the first term (a = 7) and the common difference (d = 6) from the A.P. Then, substitute these values into the arithmetic progression formula t_n = a + (n-1)d with n=19.

🎯 Exam Tip: Double-check your calculation of the common difference 'd'. A small error here will lead to an incorrect final answer for the nth term. Be careful with multiplication and addition steps.

Question 5. Find the 27th term of the following A.P. 9,4,-1,-6,-11,...
Answer:
Solution:
The given A.P. is 9, 4, -1, -6, -11,...
Here, a = 9, d = 4 - 9 = -5
Since, t_n = a + (n - 1)d
\( \implies \) t₂₇ = 9 + (27 - 1)(-5)
= 9 + 26 × (-5)
= 9 - 130
\( \implies \) t₂₇ = -121
\( \implies \) 27th term of the given A.P. is -121.
In simple words: For this A.P., the first term 'a' is 9 and the common difference 'd' is -5. Using the formula t_n = a + (n-1)d, substitute these values with n=27 to find the 27th term.

🎯 Exam Tip: Pay close attention to negative signs when calculating the common difference and performing multiplications, especially in sequences with decreasing terms. Keep track of operations with positive and negative numbers carefully.

Question 6. Find how many three digit natural numbers are divisible by 5.
Answer:
Solution:
The three digit natural numbers divisible by
5 are 100, 105, 110, ...,995
The above sequence is an A.P.
\( \implies \) a = 100, d = 105 - 100 = 5
Let the number of terms in the A.P. be n.
Then, t_n = 995
Since, t_n = a + (n - 1)d
\( \implies \) 995 = 100 + (n - 1)5
\( \implies \) 995 - 100 = (n - 1)5
\( \implies \) 895 = (n - 1)5

\( \implies \) n - 1 = \( \frac{895}{5} \)
\( \implies \) n - 1 = 179
\( \implies \) n = 179 + 1 = 180
\( \implies \) There are 180 three digit natural numbers which are divisible by 5.
In simple words: First, identify the smallest and largest three-digit numbers divisible by 5. These form an A.P. with a = 100, d = 5, and t_n = 995. Use the formula t_n = a + (n-1)d to solve for 'n', which represents the count of such numbers.

🎯 Exam Tip: Correctly identifying the first term 'a' and the last term 't_n' in the range is vital. Make sure to perform the algebraic manipulation accurately to solve for 'n', especially when dividing and adding/subtracting 1.

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.
Answer:
Solution:
For an A.P., let a be the first term and d be the common difference,
t₁₁ = 16, t₂₁ = 29 ...[Given]
t_n = a + (n - 1)d
\( \implies \) t₁₁ = a + (11 - 1)d
\( \implies \) 16 = a + 10d
i.e. a + 10d = 16 ...(i)
Also, t₂₁ = a + (21 - 1)d
\( \implies \) 29 = a + 20d
i.e. a + 20d = 29 ...(ii)
Subtracting equation (i) from (ii), we get a
a + 20d = 29
a + 10d = 16
--------------
10d = 13

\( \implies \) d = \( \frac{13}{10} \)
Substituting d = \( \frac{13}{10} \) in equation (i), we get
a + 10\( \left(\frac{13}{10}\right) \) = 16
a + 13 = 16
a = 16 - 13 = 3
t₄₁ = 3 + (41 - 1) \( \left(\frac{13}{10}\right) \)
= 3 + 40 x \( \frac{13}{10} \)
= 3 + 52
\( \implies \) t₄₁ = 55
\( \implies \) 41st term of the A.P. is 55.
In simple words: Use the given terms (t₁₁=16, t₂₁=29) and the A.P. formula t_n = a + (n-1)d to form two linear equations in 'a' and 'd'. Solve these simultaneous equations to find 'a' and 'd', then use these values to calculate the 41st term.

🎯 Exam Tip: This problem requires solving simultaneous equations. Ensure you accurately substitute the values of 'n' into the general formula to get correct equations. Be meticulous with algebraic operations, especially when dealing with fractions, to avoid calculation errors.

Question 8. 11, 8, 5, 2, ... In this A.P. which term is number-151?
Answer:
Solution:
The given A.P. is 11, 8, 5, 2,...
Here, a = 11, d = 8 - 11 = -3
Let the n^th term of the given A.P. be -151.
Then, t_n = -151
Since, t_n = a + (n - 1)d
\( \implies \) -151= 11 + (n - 1)(-3)
\( \implies \) -151 - 11 = (n - 1)(-3)
\( \implies \) -162 = (n - 1)(-3)

\( \implies \) n - 1 = \( \frac{-162}{-3} \)
\( \implies \) n - 1 = 54
\( \implies \) n = 54 + 1 = 55
\( \implies \) 55th term of the given A.P. is -151.
In simple words: To find which term is -151, first identify the first term 'a' (11) and common difference 'd' (-3). Set t_n = -151 in the formula t_n = a + (n-1)d and solve for 'n'.

🎯 Exam Tip: Pay close attention to the signs of numbers, especially when 'd' is negative. Carefully isolate 'n' by performing inverse operations step-by-step. A positive integer value for 'n' confirms that the number belongs to the A.P.

Question 9. In the natural numbers from 10 to 250, how many are divisible by 4?
Answer:
Solution:
The natural numbers from 10 to 250 divisible
by 4 are 12, 16, 20, ...,248
The above sequence is an A.P.
\( \implies \) a = 12, d = 16 - 12 = 4
Let the number of terms in the A.P. be n.
Then, t_n = 248
Since, t_n = a + (n - 1)d
\( \implies \) 248 = 12 + (n - 1)4
\( \implies \) 248 - 12 = (n - 1)4
\( \implies \) 236 = (n - 1)4

\( \implies \) n - 1 = \( \frac{236}{4} \)
\( \implies \) n - 1 = 59
\( \implies \) n = 59 + 1 = 60
\( \implies \) There are 60 natural numbers from 10 to 250 which are divisible by 4.
In simple words: Find the first number in the range [10, 250] divisible by 4 (which is 12) and the last (which is 248). These form an A.P. with a = 12 and d = 4. Use t_n = a + (n-1)d with t_n = 248 to calculate 'n', the count of such numbers.

🎯 Exam Tip: Correctly identify the first and last terms of the A.P. within the given range. Be precise with the algebraic steps to solve for 'n', as a calculation error will lead to an incorrect count.

Question 10. In an A.P. 17th term is 7 more than its 10th term. Find the common difference.
Answer:
Solution:
For an A.P., let a be the first term and d be the common difference.
According to the given condition,
t₁₇ = t₁₀ + 7
\( \implies \) a + (17-1)d = a + (10 - 1)d + 7 ...[: t_n = a + (n - 1)d]
\( \implies \) a + 16d = a + 9d + 7
\( \implies \) a + 16d - a - 9d = 7
\( \implies \) 7d = 7

\( \implies \) d = \( \frac{7}{7} \)
= 1
\( \implies \) The common difference is 1.
In simple words: Translate the given condition (17th term is 7 more than 10th term) into an algebraic equation using the A.P. formula t_n = a + (n-1)d. Simplify the equation to solve for the common difference 'd'.

🎯 Exam Tip: This problem tests your understanding of forming equations from verbal statements about A.P. terms. Remember that 'a' will cancel out, simplifying the equation significantly. Accurate application of the t_n formula is key.

Question 1. 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! She was shocked to find it. You too assume that Kabir grows in A.P. and find out his height at the age of 15 years. (Textbook pg. no. 63)
Answer:
Solution:
Height of Kabir when he was 1 year old = 70 cm Height of Kabir when he was 2 years old = 80 cm
Height of Kabir when he was 3 years old = 90 cm The heights of Kabir form an A.P.
Here, a = 70, d = 80 - 70 = 10
We have to find height of Kabir at the age of 15years i.e. t₁₅.
Now, t_n = a + (n - 1)d
\( \implies \) t₁₅ = 70 + (15 - 1)10
= 70 + 14 × 10 = 70 + 140
\( \implies \) t₁₅ = 210
\( \implies \) The height of Kabir at the age of 15 years will be 210 cm.
In simple words: The heights form an A.P. where 'a' is 70 cm (at 1 year) and 'd' is 10 cm (growth per year). To find his height at 15 years (t₁₅), apply the A.P. formula t_n = a + (n-1)d with n=15.

🎯 Exam Tip: Clearly define 'a' as the height at 1 year (t₁) and 'd' as the annual growth. Ensure 'n' correctly represents the age for the desired term. This problem combines real-world context with A.P. calculations.

Question 2. 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.? (Textbook pg. no, 62)
Answer:
Solution:
i. The given sequence is
5, 8,11,14,...
Here, t₁ = 5, t₂ = 8, t₃ = 11, t₄ = 14
\( \implies \) t₂-t₁ = 8-5 = 3
t₃-t₂ = 11-8 = 3
t₄-t₃ = 14-11 = 3
\( \implies \) t₂ - t₁ = t₃ - t₂ = t₄ - t₃ = 3 = d = constant
The difference between two consecutive terms is constant
\( \implies \) The given sequence is an A.P.
ii. t_n = a + (n - 1)d
\( \implies \) t₁₀₀ = 5 + (100 - 1)3 ...[: a = 5, d = 3]
= 5 + 99 × 3
= 5 + 297
\( \implies \) t₁₀₀ = 302
\( \implies \) 100th term of the given A.P. is 302.
iii. To check whether 92 is in given A.P., let t_n = 92
\( \implies \) t_n = a + (n - 1)d
\( \implies \) 92 = 5 + (n - 1)3
\( \implies \) 92 = 5 + 3n - 3
\( \implies \) 92 = 2 + 3n
\( \implies \) 90 = 3n

\( \implies \) n = \( \frac{90}{3} \)
= 30
\( \implies \) 92 is the 30th term of given A.P.
iv. To check whether 61 is in given A.P., let t_n = 61
61 = 5 + (n - 1)3
\( \implies \) 61 = 5 + 3n - 3
\( \implies \) 61 = 2 + 3n
\( \implies \) 61 - 2 = 3n
\( \implies \) 59 = 3n

\( \implies \) n = \( \frac{59}{3} \)
But, n is natural number

\( \implies \) n \( \ne \frac{59}{3} \)
\( \implies \) 61 is not in given A.P.
In simple words: First, confirm if the sequence is an A.P. by checking the common difference. Then, use t_n = a + (n-1)d to find the 100th term. To check if a number (like 92 or 61) is in the A.P., set t_n to that number and solve for 'n'. If 'n' is a positive integer, the number belongs to the A.P.; otherwise, it does not.

🎯 Exam Tip: When solving for 'n' to check if a number belongs to an A.P., a non-integer or negative value for 'n' indicates the number is not part of the sequence. Be thorough in evaluating all parts of such multi-faceted questions.