Class 10th Maths - Arithmetic Progressions Case Study Questions and Answers 2022 - 2023
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Arithmetic Progressions Case Study Questions With Answer Key
10th Standard CBSE
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Maths
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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 7th sample from the last?
(a) 952 (b) 945 (c) 959 (d) 966 (v) The number of bacteria in 50th sample is
(a) 546 (b) 553 (c) 448 (d) 496 -
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 -
Meenas mother start a new shoe shop. To display the shoes, she put 3 pairs of shoes in 1st row,S pairs in 2nd row, 7 pairs in 3rd row and so on.
On the basis of above information, answer the following questions.
(i) If she puts a total of 120 pairs of shoes, then the number of rows required are(a) 5 (b) 6 (c) 7 (d) 10 (ii) Difference of pairs of shoes in 17th row and 10th row is
(a) 7 (b) 14 (c) 21 (d) 28 (iii) On next day, she arranges x pairs of shoes in 15 rows, then x =
(a) 21 (b) 26 (c) 31 (d) 42 (iv) Find the pairs of shoes in 30th row.
(a) 61 (b) 67 (c) 56 (d) 59 (v) The total number of pairs of shoes in 5th and 8th row is
(a) 7 (b) 14 (c) 28 (d) 56 -
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 -
In a board game, the number of sea shells in various cells forms an A.P. If the number of sea shells in the 3rd and 11th cell together is 68 and number of shells in 11th cell is 24 more than that of 3rd cell, then answer the following
questions based on this data.
(i) What is the difference between the number of sea shells in the 19th and 20th cells?(a) 2 (b) 3 (c) 8 (d) 7 (ii) How many sea shells are there in the first cell?
(a) 52 (b) 18 (c) 16 (d) 54 (iii) How many total sea shells are there in first 13 cells?
(a) 442 (b) 221 (c) 204 (d) Can't be determined (iv) Altogether, how many sea shells are there in the first 5 cells?
(a) 220 (b) 125 (c) 96 (d) 110 (v) What is the sum of number of sea shells in the 7th and 9th cell?
(a) 42 (b) 32 (c) 74 (d) 80 -
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 -
While playing a treasure hunt game, some clues (numbers) are hidden in various spots collectively forms an A.P. If the number on the nth spot is 20 + 4n, then answer the following questions to help the player in spotting the clues.
(i) Which number is on the first spot?(a) 20 (b) 24 (c) 16 (d) 28 (ii) Which number is on the (n - 2)th spot?
(a) 16+4n (b) 24+4n (c) 12+4n (d) 28+4n (iii) Which number is on the 34th spot?
(a) 156 (b) 116 (c) 120 (d) 160 (iv) What is the sum of all the numbers on the first 10 spots?
(a) 410 (b) 420 (c) 480 (d) 410 (v) Which spot is numbered as 116?
(a) 5th (b) 8th (c) 9th (d) 24th -
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 -
Do you know, we can find A.P. in many situations in our day-to-day life. One such example is a tissue paper roll, in which the first term is the diameter of the core of the roll and twice the thickness of the paper is the common difference. If the sum of first n rolls of tissue on a roll is Sn = 0.1 n2 +7.9n, then answer the following questions.
(i) Find Sn - 1·(a) 0.1n2 - 0.2n - 7.8 (b) 0.1n2 - 7.9n (c) 0.1n2 + 7.7n - 7.8 (d) None of these (ii) Find the radius of the core.
(a) 8 cm (b) 4 cm (c) 16 cm (d) Can't be determined (iii) S2 =
(a) 16.2 (b) 8.2 (c) 2.8 (d) 4.8 (iv) What is the diameter of roll when one tissue sheet is rolled over it?
(a) 7.6 cm (b) 7.9 cm (c) 8.1 cm (d) 8.2 cm (v) Find the thickness of each tissue sheet
(a) 2 cm (b) 1 cm (c) 1 mm (d) 2 mm -
The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.
(i) Find the production during first year.(a) Rs. 5000 (b) Rs. 2200 (c) Rs. 10000 (d) none of these (ii) Find the production during 8th year
(a) Rs. 7200 (b) Rs. 22000 (c) Rs. 20400 (d) none of these (iii) Find the production during first 3 years.
(a) Rs. 21600 (b) Rs. 22000 (c) Rs. 20400 (d) none of these (iv) In which year, the production is Rs. 29,200.
(a) 10 (b) 11 (c) 12 (d) 13 (v) Find the difference of the production during 7th year and 4th year.
(a) Rs. 5000 (b) Rs. 2200 (c) Rs. 10000 (d) none of these -
(a) Find the total number of rows of candies.(i) 12 (ii) 10 (iii) 14 (iv) 8 (b) How many candies are placed in last row?
(i) 22 (ii) 21 (iii) 24 (iv) 18 (c) Find the difference in number of candies placed in 7th and 3rd row.
(i) 8 (ii) 10 (iii) 12 (iv) 14 (d) If Aditya decides to make 15 rows, then how many total candies will be placed by him with the same arrangement?
(i) 200 (ii) 150 (iii) 255 (iv) 210 (e) Find the number of candies in 12th row.
(i) 21 (ii) 30 (iii) 25 (iv) 19 -
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that each section of each class would plant twice as many plants as the class standard. There were 3 sections of each standard from 1 to 12. So, if there are three sections in class 1 say 1A, 1B and 1C, then each section would plant 2 trees. Similarly, each section of class 2 would plant 4 trees and so on. Thus, the number of trees planted by classes 1 to 12 formed an AP given by 6, 12, 18,...
(a) What is the common difference of the AP formed(i) 6 (ii) 5 (iii) 3 (iv) 2 (b) What will be the nth term of the AP formed?
(i) 5n (ii) 6n (iii) 5n+6 (iv) 6n+6 (c) How many trees will be planted by the students of all the sections of class 8?
(i) 42 (ii) 48 (iii) 54 (iv) 60 (d) Find the total number of trees planted by class 12 students.
(i) 54 (ii) 72 (iii) 66 (iv) None of these (e) What will be the third term from the end of the AP formed?
(i) 72 (ii) 66 (iii) 60 (iv) 54
Case Study
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Answers
Arithmetic Progressions Case Study Questions With Answer Key Answer Keys
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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 = 122th term = 105 + 121(7) = 952
(v) (c): t50 = 105 + 49 x 7 = 448 -
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 -
Number of pairs of shoes in 1st, 2nd, 3rd row, ... are 3,5,7, ...
So, it forms an A.P. with first term a = 3, d = 5 - 3 = 2
(i) (d): Let n be the number of rows required.
\(\therefore S_{n}=120 \)
\(\Rightarrow \quad \frac{n}{2}[2(3)+(n-1) 2]=120 \)
\(\Rightarrow \quad n^{2}+2 n-120=0 \Rightarrow n^{2}+12 n-10 n-120=0\)
\(\Rightarrow \quad(n+12)(n-10)=0 \Rightarrow n=10\)
So, 10 rows required to put 120 pairs.
(ii) (b): No. of pairs in 1ih row = t17 = 3 + 16(2) = 35
No. of pairs in 10th row = t10 = 3 + 9(2) = 21
\(\therefore\) Required difference = 35 - 21 = 14
(iii) (c) : Here n = 15
\(\therefore\) t15 = 3 + 14(2) = 3 + 28 = 31
(iv) (a): No. of pairs in 30th row = t30 = 3 +29(2) = 61
(v) (c): No. of pairs in 5th row = t5 = 3 + 4(2) = 11
No. of pairs in 8th row = t8 = 3 + 7(2) = 17
\(\therefore\) Required sum = 11 + 17 = 28 -
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 number of sea shells in the cells be of the form a, a + d, a + 2d, ...
According to question, we have
(a + 2d) + (a + 10d) = 68
\(\Rightarrow\) 2a + 12d = 68 \(\Rightarrow\) a + 6d = 34 ... (i)
Also, (a + 10d) - (a + 2d) = 24 \(\Rightarrow\) d = 3
From (1), we get a + 6(3) = 34 \(\Rightarrow\) a = 16
(i) (b): Required difference, d = 3
(ii) (c): Number of sea shells in the first cell (a) = 16
(iii) (a): Total number of sea shells in 13 cells = S13
\(=\frac{13}{2}[2(16)+12(3)]=6.5(68)=442\)
(iv) (d): \(S_{5}=\frac{5}{2}[2(16)+4(3)]=110\)
(v) (c): Required sum = t7 + t9 = (a + 6d) + (a + 8d)
= 2(16) + 14(3) = 74 -
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): \(S_{15}=\frac{15}{2}[2(7)+14(-5)]=-420\) -
Number on nth spot = 20 + 4n i.e., tn = 20 + 4n
(i) (b): Number on 1st spot = t1 = 20 + 4(1) = 24
(ii) (c): Number on (n - 2)th spot = tn - 2
= 20 + 4 (n - 2)
= 20 + 4n - 8 = 12 + 4n
(iii) (a): Number on 34th spot = t34 = 20 + 4(34) = 156
(iv) (b): Here a = t1 = 24
Now, t2 = 20 + 4 (2) = 20 + 8 = 28
\(\therefore\) d = t2 - t1 = 4
So, required sum \(=S_{10}=\frac{10}{2}[2(24)+9(4)]=420\)
(v) (d): Let nth spot is numbered as 116.
\(\therefore\) tn = 116
\(\Rightarrow 20+4 n=116 \Rightarrow 4 n=96 \Rightarrow n=24\) -
(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 -
Here Sn = 0.1n2 + 7.9n
(i) (c): Sn -1= 0.1(n - 1)2 + 7.9(n - 1)
= 0.1n2 + 7.7n - 7.8
(ii) (b): S1 = t1 = a = 0.1(1)2 + 7.9(1) = 8 cm = Diameter of core
So, radius of the core = 4 cm
(iii) (a): S2 = 0.1(2)2 + 7.9(2) = 16.2
(iv) (d): Required diameter = t2 = S2 - S1 = 16.2 - 8 = 8.2 cm
(v) (c): As d = t2 - t1 = 8.2 - 8 = 0.2 cm
So, thickness of tissue = 0.2 \(\div\) 2 = 0.1 cm = 1 mm -
(i) (a): Let the production during first year be a and let d be the increase in production every year. Then,
a6 = 16000 \(\Rightarrow\)a + 5d = 16,000 (i)
and a9 = 22600 \(\Rightarrow\)a + 8d = 22600 (ii)
On substracting (i) from (ii) , we get
3d = 6600 \(\Rightarrow\) d = 2200
Putting d = 2200 in (i) we get,
a + 5 x 2200 = 16000
\(\Rightarrow\) a + 11000 = 16000 \(\Rightarrow\) a = 16000 - 11000 = 5000
Thus , a = 5000 and d = 2200
Production during first year, a = 5000.
(ii) (c): Production during 8th year is given by a8
= (a + 7d)
= (5000 + 7(2200))
= (5000 + 15400)
= 20400.
(iii) (a): a2 = (a + d)
= (5000 + 2200) = 7200.
a3 = (a2 + d)
= 7200 + 2200 = 9400.
Production during first 3 years = 5000 + 7200 + 9400 = 21600
(iv) (c): an = 5000 + (n – 1)2200 = 29200
(n – 1)2200 = 29200 – 5000 = 24200
⇒ n – 1 = 11
⇒ n = 12
(v) (d): a4 = (a + 3d)
= (5000 + 3(2200)) = 5000 + 6600 = 11600.
a7 = (a6 + d)
= 16000 + 2200 = 18200.
Difference = 18200 – 11600 = 6600 -
(a) (ii) 10
(b) (ii) 21
(c) (i) 8
(d) (iii) 255
(e) (iii) 25 -
(a) (i)
The given AP is 6,12,18...,
The common difference = 12-6 =6.
(b) (ii)
In the given AP, we have:a=d=6
\(\therefore a_{n}=a+(n-1) d=6+(n-1) 6=6+6 n-6=6 n\)
(c) (ii)
The number of trees planted by the students of all the sections of class
= 8th term of the given AP
= 6n = 6X8=48
(d) (ii) T
otal number of trees planted by class 12 students
= 6 X 12 = 72
(e) (iii)
3rd term from the end = \((n-3+1) \text { th term }\)
= (12-3+1) th term = 10th term
= 6X10=60
