CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set H

TOPICS COVERED

• 1. Sequence / Progression

• 2. Arithmetic Progressions and its \( n^{th} \) term
• 3. Sum of First \( n \) Terms of an AP

1. SEQUENCE/PROGRESSION

• Sequence/Progression: A sequence/progression is a succession of numbers or terms formed according to some pattern or rule. Various numbers occurring in a sequence are called terms or elements.

Consider the following arrangements of numbers:
(i) 1, 8, 27, 64, 125, ...
(ii) 1, \( \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \dots \)
(iii) 2, 4, 6, 8, 10, ...

In each of the above arrangements, numbers are arranged in a definite order according to some rule. So, they are sequences.

A sequence is generally written as \( < a_n > : a_1, a_2, a_3, \dots, a_n \) where \( a_1, a_2, a_3, \dots \) are the first, second and third terms of the sequence.

• A sequence with finite number of terms or numbers is called a finite sequence.
• A sequence with infinite number of terms or numbers is called an infinite sequence.

Question. Write first four terms of each of the following sequence, whose general terms are: (i) \( a_n = 3n - 7 \) (ii) \( a_n = (-1)^{n+1} \times 3^n \)
Answer: (i) \( a_n = 3n - 7 \)
\( a_1 = 3 \times 1 - 7 = 3 - 7 = -4 \),
\( a_2 = 3 \times 2 - 7 = 6 - 7 = -1 \),
\( a_3 = 3 \times 3 - 7 = 9 - 7 = 2 \), and
\( a_4 = 3 \times 4 - 7 = 12 - 7 = 5 \)
(ii) \( a_n = (-1)^{n+1} \times 3^n \)
\( a_1 = (-1)^{1+1} \times 3^1 = 3 \),
\( a_2 = (-1)^{2+1} \times 3^2 = (-1)^3 \times 3^2 = -9 \),
\( a_3 = (-1)^4 \times 3^3 = 27 \) and \( a_4 = (-1)^5 \times 3^4 = -81 \)

Question. What is \( 18^{th} \) term of the sequence defined by \( a_n = \frac{n(n - 3)}{n + 4} \)?
Answer: We have, \( a_n = \frac{n(n - 3)}{n + 4} \)
Putting \( n = 18 \), we get \( a_{18} = \frac{18 \times (18 - 3)}{18 + 4} \)
\( = \frac{18 \times 15}{22} = \frac{135}{11} \)

Exercise 

Multiple Choice Questions (MCQs)

Question. If \( a_n = 5n - 4 \) is a sequence, then \( a_{12} \) is
(a) 48
(b) 52
(c) 56
(d) 62
Answer: (c) 56

Question. If \( a_n = 3n - 2 \), then the value of \( a_7 + a_8 \) is
(a) 39
(b) 41
(c) 47
(d) 53
Answer: (b) 41

Question. The second term of the sequence defined by \( a_n = 3n + 2 \) is
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (d) 8

Assertion-Reason Type Questions

Question. Assertion (A): The arrangement of numbers, i.e., – 4, 16, – 64, 256, – 1024, 4096, ... form a sequence.
Reason (R): An arrangement of numbers which are arranged in a definite order according to some rule, is called a sequence.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question. Assertion (A): Sequence 1, 5, 9, 13, 17, 21, ... is a finite sequence.
Reason (R): A sequence with finite number of terms or numbers is called a finite sequence.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (d) Assertion (A) is false but reason (R) is true.

2. ARITHMETIC PROGRESSION AND ITS \( n^{th} \) TERM

• An arithmetic progression is a sequence of numbers in which each term is obtained by adding a fixed number ‘d’ to the preceding term, except the first term ‘a’. This fixed number is known as common difference of the AP. Common difference of an AP can be negative, positive or zero.

The general form of an AP is \( a, a + d, a + 2d, a + 3d, \dots \)

Examples:
(i) The sequence 1, 4, 7, 10, 13, ... is an AP whose first term is 1 and the common difference is equal to 3.
(ii) The sequence 11, 7, 3, –1, ... is an AP whose first term is 11 and the common difference is equal to –4.

• In the list of numbers \( a_1, a_2, a_3, \dots \) if the differences \( a_2 – a_1, a_3 – a_2, a_4 – a_3, \dots \) give the same value, i.e., if \( a_{k+1} – a_k \) is the same for different values of \( k \), then the given list of numbers is an AP.
• The \( n^{th} \) term \( a_n \) (or the general term) of an AP is \( a_n = a + (n – 1) d \), where \( a \) is the first term, \( d \) is the common difference and \( n \) is the number of terms. Also, \( d = a_{n+1} – a_n \).
• If three terms \( a, b \) and \( c \) are in AP, then \( b – a = c – b \) or \( 2b = a + c \).
• If \( l \) is the last term of an AP, then \( n^{th} \) term from the end of the AP = \( l + (n – 1)(-d) = l – (n – 1)d \).

Question. Is 0 a term of the AP: 31, 28, 25, ...? Justify your answer.
Answer: Given AP is 31, 28, 25, ...
Here, \( a = 31 \), \( d = 28 – 31 = – 3 \)
For 0 be a term of this AP, \( 0 = a_n \) for some \( n \)
\( \Rightarrow 0 = a + (n – 1)d \)
\( \Rightarrow 0 = 31 + (n – 1) (– 3) \Rightarrow 31 – 3n + 3 = 0 \)
\( \Rightarrow – 3n = – 34 \Rightarrow n = \frac{34}{3} = 11 \frac{1}{3} \)
which is not possible as \( n \) cannot be a fraction.
Therefore, 0 cannot be a term of this AP.

Question. Find the \( 12^{th} \) term from the end of the AP: –2, – 4, – 6, ..., – 100.
Answer: Let \( a \) be the first term, \( d \) the common difference and \( l \) the last term of AP.
Here, \( a = –2 \), \( d = (–4 + 2) = –2 \), \( l = –100 \) and \( n = 12 \)
\( n^{th} \) term from end = \( l – (n – 1)d \)
\( 12^{th} \) term from end = \( –100 – (12 – 1) (–2) = –100 + 22 = -78 \)

Question. For what value of \( x \): \( 2x \), \( x + 10 \) and \( 3x + 2 \) are in AP?
Answer: Since, given numbers are in AP.
So, \( (x + 10) – 2x = (3x + 2) – (x + 10) \)
\( \Rightarrow –x + 10 = 2x – 8 \)
\( \Rightarrow 3x = 18 \) or \( x = 6 \)

Question. Find the \( 25^{th} \) term of the AP: \( -5, \frac{-5}{2}, 0, \frac{5}{2}, \dots \)
Answer: We have, \( a = –5 \), \( d = \frac{-5}{2} – (–5) = \frac{5}{2} \)
\( a_n = a + (n – 1) d \)
\( a_{25} = (–5) + (25 – 1) \frac{5}{2} = (–5) + 24 \left( \frac{5}{2} \right) = -5 + 60 = 55 \)

Question. Find the \( 20^{th} \) term from the last term of the AP: 3, 8, 13,..., 253.
Answer: Given, last term = \( l = 253 \)
And, common difference = \( d = 8 – 3 = 5 \)
\( 20^{th} \) term from end = \( l – (n – 1) \times d = 253 – 19 \times 5 = 253 – 95 = 158 \)

Question. Which term of the AP: 3, 8, 13, 18, ..., is 78?
Answer: Let \( a_n \) be the required term of the AP: 3, 8, 13, 18,...
Here, \( a = 3 \), \( d = 8 – 3 = 5 \) and \( a_n = 78 \)
Now, \( a_n = a + (n – 1)d \)
\( \Rightarrow 78 = 3 + (n – 1) \times 5 \Rightarrow 78 – 3 = (n – 1) \times 5 \)
\( \Rightarrow 75 = (n – 1) \times 5 \Rightarrow \frac{75}{5} = n – 1 \)
\( \Rightarrow 15 = n – 1 \Rightarrow n = 15 + 1 = 16 \)
Hence, \( 16^{th} \) term of given AP is 78.

Question. The sum of the \( 5^{th} \) and \( 7^{th} \) terms of an AP is 52 and the \( 10^{th} \) term is 46. Find the AP.
Answer: Let the first term and the common difference of an AP be ‘\( a \)’ and ‘\( d \)’.
\( a_5 = a + 4d \) and \( a_7 = a + 6d \)
So, \( a_5 + a_7 = 2a + 10d = 52 \Rightarrow 2a + 10d = 52 \dots (i) \)
Also, \( a_{10} = a + 9d = 46 \Rightarrow a + 9d = 46 \dots (ii) \)
From (i) and (ii), \( d = 5 \) and \( a = 1 \)
So, the AP is as follows 1, 6, 11, 16, 21, ...

Question. An AP consists of 50 terms of which \( 3^{rd} \) term is 12 and the last term is 106. Find the \( 29^{th} \) term.
Answer: Let \( a \) be the first term and \( d \) be the common difference.
Since, given AP has 50 terms, so \( n = 50 \)
\( a_3 = 12 \Rightarrow a + (3 – 1)d = 12 \)
\( \Rightarrow a + 2d = 12 \dots (i) \)
Also, \( a_{50} = 106 \Rightarrow a + (50 – 1)d = 106 \)
\( \Rightarrow a + 49d = 106 \dots (ii) \)
Subtracting (i) from (ii), we get
\( 47d = 94 \Rightarrow d = \frac{94}{47} = 2 \)
Putting the value of \( d \) in equation (i), we get
\( a + 2 \times 2 = 12 \Rightarrow a = 12 – 4 = 8 \)
Here, \( a = 8 \), \( d = 2 \)
So, \( 29^{th} \) term of the AP is given by
\( a_{29} = a + (29 – 1)d = 8 + 28 \times 2 \Rightarrow a_{29} = 8 + 56 \Rightarrow a_{29} = 64 \)

Question. The first term of an AP is \( x \) and its common difference is \( y \). Find its \( 12^{th} \) term.
Answer: \( a_{12} = a + 11d = x + 11y \).

Exercise 

Multiple Choice Questions (MCQs)

Question. In an AP, if \( d = – 4 \), \( n = 7 \), \( a_n = 4 \), then \( a \) is
(a) 6
(b) 7
(c) 20
(d) 28
Answer: (d) 28

Question. The \( n^{th} \) term of the AP: \( a, 3a, 5a, \dots \) is
(a) \( na \)
(b) \( (2n – 1)a \)
(c) \( (2n + 1)a \)
(d) \( 2na \)
Answer: (b) (2n – 1)a

Question. The first term of an AP is \( p \) and the common difference is \( q \), then its \( 10^{th} \) term is
(a) \( q + 9p \)
(b) \( p – 9q \)
(c) \( p + 9q \)
(d) \( 2p + 9p \)
Answer: (c) p + 9q

Question. If \( \frac{4}{5}, a, 2 \) are three consecutive terms of an AP, then the value of \( a \) is
(a) \( \frac{5}{2} \)
(b) \( \frac{2}{7} \)
(c) \( \frac{5}{7} \)
(d) \( \frac{7}{5} \)
Answer: (d) \frac{7}{5}

Assertion-Reason Type Questions

Question. Assertion (A): Common difference of the AP: \( –5, –1, 3, 7, \dots \) is 4.
Reason (R): Common difference of the AP : \( a, a + d, a + 2d, \dots \) is given by \( d = 2^{nd} \) term \( – 1^{st} \) term.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question. Assertion (A): If \( n^{th} \) term of an AP is \( 7 – 4n \), then its common difference is \( – 4 \).
Reason (R): Common difference of an AP is given by \( d = a_{n+1} – a_n \).
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Question. Assertion (A): Common difference of an AP in which \( a_{21} – a_7 = 84 \) is 14.
Reason (R): \( n^{th} \) term of an AP is given by \( a_n = a + (n – 1) d \).
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer: (d) Assertion (A) is false but reason (R) is true.

Case Study Based Questions

Your friend Veer wants to participate in a 200 m race. Presently, he can run 200 m in 51 seconds and during each day practice it takes him 2 seconds less. He wants to do in 31 seconds.

Question. Which of the following terms are in AP for the given situation?
(a) 51, 53, 55, ...
(b) 51, 49, 47, ...
(c) –51, –53, –55, ...
(d) 51, 55, 59, ...
Answer: (b) 51, 49, 47, ...

Question. What is the minimum number of days he needs to practice till his goal is achieved?
(a) 10
(b) 12
(c) 11
(d) 9
Answer: (c) 11

Question. Which of the following term is not in the AP of the above given situation?
(a) 41
(b) 30
(c) 37
(d) 39
Answer: (b) 30

Question. If \( n^{th} \) term of an AP is given by \( a_n = 2n + 3 \) then common difference of an AP is
(a) 2
(b) 3
(c) 5
(d) 1
Answer: (a) 2

Question. The value of \( x \), for which \( 2x, x + 10, 3x + 2 \) are three consecutive terms of an AP is
(a) 6
(b) – 6
(c) 18
(d) –18
Answer: (a) 6

India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in \( 6^{th} \) year and 22600 in \( 9^{th} \) year.

Question. The production during first year is
(a) 3000 TV sets
(b) 5000 TV sets
(c) 7000 TV sets
(d) 10000 TV sets
Answer: (b) 5000 TV sets

Question. The production during \( 8^{th} \) year is
(a) 10500
(b) 11900
(c) 12500
(d) 20400
Answer: (d) 20400

Question. The production during first 3 years is
(a) 12800
(b) 19300
(c) 21600
(d) 25200
Answer: (c) 21600

Question. In which year, the production is 29,200?
(a) \( 10^{th} \) year
(b) \( 12^{th} \) year
(c) \( 15^{th} \) year
(d) \( 18^{th} \) year
Answer: (b) 12th year

Question. The difference of the production during \( 7^{th} \) year and \( 4^{th} \) year is
(a) 6600
(b) 6800
(c) 5400
(d) 7200
Answer: (a) 6600