CBSE 10th Standard Maths Subject Arithmetic Progressions Case Study Questions With Solution 2021
QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get more marks in Exams
QB365 - Question Bank Software
CBSE 10th Standard Maths Subject Arithmetic Progressions Case Study Questions With Solution 2021
10th Standard CBSE
-
Reg.No. :
Maths
-
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 34th 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 19th 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 -
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 14th 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 30th 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 1st 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) 9th (b) 99th (c) 10th (d) 100th -
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 6th and 14thcard and finds their sum to be -76. On the second turn he picks up 8th and 16thcard 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 19th card?
(a) -88 (b) -82 (c) -92 (d) -102 (iv) What is the number on the 23rd card?
(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 -
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 ..... , an are in A.P., then which of the following is true?
(a) a1 + k, a2 + k, a3 + k, , an + k are in A.P., where k is a constant. (b) k - a1 k - a2, k - a3, , k - an are in A.P., where k is a constant. (c) ka1, ka2, ka3 ..... , kan are in A.P., where k is a constant. (d) All of these (iv) If the nth 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 -
Jack is much worried about his upcoming assessment on A.P. He was vigorously practicing for the exam but unable to solve some questions. One of these questions is as shown below. If the 3rd and the 9th terms of an A.P. are 4 and - 8 respectively, then help Jack in solving the problem.
(i) What is the common difference?(a) 2 (b) -1 (c) -2 (d) 4 (ii) What is the first term?
(a) 6 (b) 2 (c) -2 (d) 8 (iii) Which term of the A.P. is -160?
(a) 80th (b) 85th (c) 81th (d) 84th (iv) Which of the following is not a term of the given A.P.?
(a) -123 (b) -100 (c) 0 (d) -200 (v) What is the 75th term of the A.P.?
(a) -140 (b) -102 (c) -150 (d) -158
Case Study Questions
*****************************************
CBSE 10th Standard Maths Subject Arithmetic Progressions Case Study Questions With Solution 2021 Answer Keys
-
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 nth 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 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 14th 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 30th 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 1st day.
tn = 10(2.75) \(\Rightarrow\) a + (n - 1)d = 27.5 \(\Rightarrow\) n = 100. -
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): \(S_{15}=\frac{15}{2}[2(7)+14(-5)]=-420\) -
(i) (c)
(ii) (c)
(iii) (d)
(iv) (b)
(v) (c) -
We have, 3rd term = 4 and 9th term = -8 i.e., a + 2d = 4 ........(i)
and a + 8d = -8 .........(ii)
Solving (1) and (2), we get
d = -2, a = 8
(i) (c)
(ii) (d)
(iii) (b): Let tn = -160 \(\Rightarrow\) a + (n - 1) d = -160
\(\Rightarrow\) 8 + (n - 1)(-2) = -160 \(\Rightarrow\) (n - 1)(-2) = -168
\(\Rightarrow\) n - 1 = 84 \(\Rightarrow\) n = 85
So, t85 = -160
(iv) (a)
(v) (a): t75 = a + 74d = 8 + 74( -2) = -140
Case Study Questions
