CBSE 10th Standard Maths Subject Arithmetic Progressions Case Study Questions 2021 | QB365

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CBSE 10th Standard Maths Subject Arithmetic Progressions Case Study Questions 2021

Published on: 22/05/2021

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Std: 10th Standard CBSE - Maths

chapter: - Case study Questions

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Questions:

Case Study Questions

5 x 4 = 20

A sequence is an ordered list of numbers. A sequence of numbers such that the difference between the consecutive terms is constant is said to be an arithmetic progression (A.P.).
On the basis of above information, answer the following questions.
(i) Which of the following sequence is an A.P.?

(a) 10,24,39,52,....

(b) 11,24,39,52, ...

(c) 10,24,38,52, ...

(d) 10, 38, 52, 66, ....

(ii) If x, y and z are in A.P., then

(a) x + z = y

(b) x - z = y

(c) x + z = 2y

(d) None of these

(iii) If a₁ a₂, a₃ ..... , a_n are in A.P., then which of the following is true?

(a) a₁ + k, a₂ + k, a₃ + k, , a_n + k are in A.P., where k is a constant.

(b) k - a₁ k - a₂, k - a₃, , k - a_n are in A.P., where k is a constant.

(c) ka₁, ka₂, ka₃ ..... , ka_n are in A.P., where k is a constant.

(d) All of these

(iv) If the n^th term (n > 1) of an A.P. is smaller than the first term, then nature of its common difference (d) is

(a) d > 0	(b) d < 0
(c) d = 0	(d) Can't be determined

(v) Which of the following is incorrect about A.P.?

(a) All the terms of constant A.P. are same.

(b) Some terms of an A.P. can be negative.

(c) All the terms of an A.P. can never be negative.

(d) None of these

Amit was playing a number card game. In the game, some number cards (having both +ve or -ve numbers) are arranged in a row such that they are following an arithmetic progression. On his first turn, Amit picks up 6^th and 14^thcard and finds their sum to be -76. On the second turn he picks up 8^th and 16^thcard and finds their sum to be -96. Based on the above information, answer the following questions.

(i) What is the difference between the numbers on any two consecutive cards?

(a) 7

(b) -5

(c) 11

(d) -3

(ii) The number on first card is

(a) 12

(b) 3

(c) 5

(d) 7

(iii) What is the number on the 19^th card?

(a) -88

(b) -82

(c) -92

(d) -102

(iv) What is the number on the 23^rdcard?

(a) -103

(b) -122

(c) -108

(d) -117

(v) The sum of numbers on the first 15 cards is

(a) -840

(b) -945

(c) -427

(d) -420

Anuj gets pocket money from his father everyday. Out of the pocket money, he saves Rs 2.75 on first day, Rs 3 on second day, Rs 3.25 on third day and so on.
On the basis of above information, answer the following questions .

(i) What is the amount saved by Anuj on 14^th day?

(a) Rs 6.25

(b) Rs 6

(c) Rs 6.50

(d) Rs 6.75

(ii) What is the total amount saved by Anuj in 8 days?

(a) Rs 18

(b) Rs 33

(c) Rs 24

(d) Rs 29

(iii) What is the amount saved by Anuj on 30^th day?

(a) Rs 10

(b) Rs 12.75

(c) Rs 10.25

(d) Rs 9.75

(iv) What is the total amount saved by him in the month of June, if he starts savings from 1^st June?

(a) Rs 191

(b) Rs 191.25

(c) Rs 192

(d) Rs 192.5

(v) On which day, he save tens times as much as he saved on day-I?

(a) 9^th

(b) 99^th

(c) 10^th

(d) 100^th

In a class the teacher asks every student to write an example of A.P. Two friends Geeta and Madhuri writes their progressions as -5, -2, 1,4, ... and 187, 184, 181, .... respectively. Now, the teacher asks various students of the class the following questions on these two progressions. Help students to find the answers of the questions.

(i) Find the 34^th term of the progression written by Madhuri.

(a) 286

(b) 88

(c) -99

(d) 190

(ii) Find the sum of common difference of the two progressions.

(a) 6

(b) -6

(c) 1

(d) 0

(iii) Find the 19^th term of the progression written by Geeta.

(a) 49

(b) 59

(c) 52

(d) 62

(iv) Find the sum of first 10 terms of the progression written by Geeta.

(a) 85

(b) 95

(c) 110

(d) 200

(v) Which term of the two progressions will have the same value?

(a) 31

(b) 33

(c) 32

(d) 30

In a pathology lab, a culture test has been conducted. In the test, the number of bacteria taken into consideration in various samples is all3-digit numbers that are divisible by 7, taken in order.

On the basis of above information, answer the following questions.
(i) How many bacteria are considered in the fifth sample?

(a) 126

(b) 140

(c) 133

(d) 149

(ii) How many samples should be taken into consideration?

(a) 129

(b) 128

(c) 130

(d) 127

(iii) Find the total number of bacteria in the first 10 samples.

(a) 1365

(b) 1335

(c) 1302

(d) 1540

(iv) How many bacteria are there in the 7^th sample from the last?

(a) 952

(b) 945

(c) 959

(d) 966

(v) The number of bacteria in 50^th sample is

(a) 546

(b) 553

(c) 448

(d) 496

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Answers Key:

Case Study Questions

5 x 4 = 20

(i) (c)
(ii) (c)
(iii) (d)
(iv) (b)
(v) (c)

Let the numbers on the cards be a, a + d, a + Zd, ...
According to question, We have (a + 5d) + (a + 13d) = -76
\(\Rightarrow\) 2a+18d = -76\(\Rightarrow\)a + 9d= -38 ... (1)
And (a + 7d) + (a + 15d) = -96
\(\Rightarrow\) 2a + 22d = -96 \(\Rightarrow\) a + 11d = -48 ...(2)
From (1) and (2), we get
2d= -10 \(\Rightarrow\) d= -5
From (1), a + 9(-5) = -38 \(\Rightarrow\) a = 7
(i) (b): The difference between the numbers on any two consecutive cards = common difference of the A.P.=-5
(ii) (d): Number on first card = a = 7
(iii) (b): Number on 19th card = a + 18d = 7 + 18(-5) = -83
(iv) (a): Number on 23rd card = a + 22d = 7 + 22( -5) = -103
(v) (d): \(S_{15}=\frac{15}{2}[2(7)+14(-5)]=-420\)

Here the savings form an A.P. i.e., Rs 2.75, Rs 3, Rs 3.25, ...
So, a = 2.75, d = 3 - 2.75 = 0.25
(i) (b): Amount saved by Anuj on 14^th day
= t₁₄ = a + 13d = 2.75 + 13(0.25) = ₹ 6
(ii) (d): Total amount saved by Anuj in 8 days
\(=S_{8}=\frac{8}{2}[2(2.75)+7(0.25)]=₹ 29\)
(iii) (a): Amount saved by Anuj on 30^th day
= t₃₀ = a + 29d = 2.75 + 29(0.25) = ₹ 10
(iv) (b): Number of days in June = 30
\(\therefore S_{30}=\frac{30}{2}[2(2.75)+29(0.25)]=₹ 191.25\)
(v) (d): Let on nth day, he saves 10 times as he saves on 1^st day.
t_n = 10(2.75) \(\Rightarrow\) a + (n - 1)d = 27.5 \(\Rightarrow\) n = 100.

Geeta's A.P. is -5, -2, 1,4, ...
Here, first term (a₁) = -5 and common difference (d₁) = -2 + 5 = 3
Similarly, Madhuri's A.P. is 187, 184, 181, ...
Here first term (a₂) = 187 and common difference (d₂) = 184 - 187 = -3
(i) (b): t₃₄ = a₂ + 33d₂ = 187 + 33(-3) = 88
(ii) (d): Required sum = 3 + (-3) = 0
(iii) (a): t₁₉ = a₁ + 18d₁ = (-5) + 18(3) = 49
(iv) (a) : \(S_{10}=\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]=\frac{10}{2}[2(-5)+9(3)]=85\)
(v) (b): Let n^th terms of the two A.P:s be equal.
\(\therefore\) -5 + (n - 1)3 = 187 + (n - 1)(-3)
\(\Rightarrow\) 6(n - 1) = 192 \(\Rightarrow\) n = 33

Here the smallest 3-digit number divisible by 7 is 105. So, the number of bacteria taken into consideration is 105, 112, 119, .... ,994 So, first term (a) = 105, d = 7 and last term = 994
(i) (c): t₅ = a + 4d = 105 + 28 = 133
(ii) (b): Let n samples be taken under consideration.
\(\because\) Last term = 994
\(\Rightarrow\) a + (n - 1)d = 994 \(\Rightarrow\) 105 + (n - 1)7 = 994 \(\Rightarrow\) n = 128
(iii) (a): Total number of bacteria in first 10 samples
\(=S_{10}=\frac{10}{2}[2(105)+9(7)]=1365\)
(iv) (a): t₇ from end = (128 - 7 + 1)^th term from beginning = 122^th term = 105 + 121(7) = 952
(v) (c): t₅₀ = 105 + 49 x 7 = 448