CBSE Class 10 Maths HOTs Arithmetic Progressions Set C

Case Based MCQs

In a flower bed, there are 43 rose plants in the first row, 41 in the second, 39 in the third and so on.

Question. If there are 11 rose plants in the last row, then number of rose required are
(a) 16
(b) 15
(c) 17
(d) 10
Answer: (c)

Question. Difference of rose plants in 7th row and 13th row is
(a) 11
(b) 12
(c) 13
(d) 14
Answer: (b)

Question. If there are x rose plants in 15 rose, then x is equal to
(a) 10
(b) 12
(c) 13
(d) 15
Answer: (d)

Question. The rose plants in 6th row is
(a) 35
(b) 37
(c) 33
(d) 31
Answer: (c)

Question. The total number of rose plants in 5th and 8th row is
(a) 64
(b) 54
(c) 46
(d) 45
Answer: (a)

The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. The sum of the first ten terms of this AP is 235.

Question. Let first term and common difference of an AP be a and d, respectively. Then pair of linear equations for given problem is
(a) \( 13a + 31d = 167, 2a + 9d = 47 \)
(b) \( 13a + 31d = 169, 2a + 9d = 45 \)
(c) \( 12a + 31d = 167, 2a + 9d = 47 \)
(d) \( 12a + 31d = 169, 2a + 9d = 45 \)
Answer: (c)

Question. Common difference of given AP is
(a) 5
(b) 7
(c) 9
(d) 11
Answer: (a)

Question. First term of given AP is
(a) 3
(b) 4
(c) 2
(d) 1
Answer: (d)

Question. Fourth term of the AP is
(a) 15
(b) 16
(c) 17
(d) 18
Answer: (b)

Question. Sum of first twenty terms is
(a) 970
(b) 990
(c) 950
(d) 980
Answer: (a)

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 6th year and 22600 in 9th year.

Question. Find the production during first year.
(a) 4000 sets
(b) 5000 sets
(c) 6000 sets
(d) 7000 sets
Answer: (b)

Question. Find the production during 8th year.
(a) 48000 sets
(b) 20400 sets
(c) 43000 sets
(d) None of these
Answer: (b)

Question. Find the production during first 3 years.
(a) 20000 sets
(b) 25000 sets
(c) 31000 sets
(d) 21600 sets
Answer: (d)

Question. In which year, the production is ₹ 29200.
(a) 11
(b) 12
(c) 10
(d) 8
Answer: (b)

Question. Find the difference of the production during 7th year and 4th year.
(a) 5500
(b) 6700
(c) 5400
(d) 6600
Answer: (d)

Your friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of 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)

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)

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)

Question. If nth 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)

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)

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of ₹ 118000 by paying every month starting with the first installment of ₹ 1000. If he increases the installment by ₹ 100 every month, answer the following :

Question. The amount paid by him in 30th installment is
(a) 3900
(b) 3500
(c) 3700
(d) 3600
Answer: (a)

Question. The amount paid by him in the 30 installments is
(a) 37000
(b) 73500
(c) 75300
(d) 75000
Answer: (b)

Question. What amount does he still have to pay after 30th installment?
(a) 45500
(b) 49000
(c) 44500
(d) 54000
Answer: (c)

Question. If total installments are 40, then amount paid in the last installment?
(a) 4900
(b) 3900
(c) 5900
(d) 9400
Answer: (a)

Question. The ratio of the 1st installment to the last installment is
(a) 1 : 49
(b) 10 : 49
(c) 10 : 39
(d) 39 : 10
Answer: (b)

Short Answer Type Questions

Question. Justify whether it is true to say that \( -1, -\frac{3}{2}, -2, -\frac{5}{2}, \dots \) forms an AP as \( a_2 - a_1 = a_3 - a_2 \).
Answer: Here, \( a_1 = -1, a_2 = -\frac{3}{2}, a_3 = -2, a_4 = -\frac{5}{2} \).
Calculating differences:
\( a_2 - a_1 = -\frac{3}{2} - (-1) = -\frac{3}{2} + 1 = -\frac{1}{2} \)
\( a_3 - a_2 = -2 - (-\frac{3}{2}) = -2 + \frac{3}{2} = -\frac{1}{2} \)
Since \( a_2 - a_1 = a_3 - a_2 \), the common difference is constant. Thus, it is true that the sequence forms an AP.

Question. Find the values of \( a, b \) and \( c \) if it is given that the numbers \( a, 7, b, 23, c \) are in AP.
Answer: Let the common difference be \( d \).
The terms are \( a, a+d, a+2d, a+3d, a+4d \).
Given: \( a+d = 7 \) and \( a+3d = 23 \).
Subtracting the equations: \( (a+3d) - (a+d) = 23 - 7 \Rightarrow 2d = 16 \Rightarrow d = 8 \).
Substituting \( d = 8 \) in \( a+d = 7 \): \( a + 8 = 7 \Rightarrow a = -1 \).
Now, \( b = a+2d = -1 + 2(8) = 15 \).
\( c = a+4d = -1 + 4(8) = 31 \).
Values are \( a = -1, b = 15, c = 31 \).

Question. The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.
Answer: Let the angles be \( a-d, a, a+d \).
Sum of angles: \( (a-d) + a + (a+d) = 180^\circ \Rightarrow 3a = 180^\circ \Rightarrow a = 60^\circ \).
Given: Greatest angle = 2 \(\times\) Least angle.
\( a+d = 2(a-d) \Rightarrow 60+d = 2(60-d) \Rightarrow 60+d = 120-2d \Rightarrow 3d = 60 \Rightarrow d = 20^\circ \).
The angles are \( 60-20, 60, 60+20 \), i.e., \( 40^\circ, 60^\circ, 80^\circ \).

Question. The taxi fare after each km, when the fare is ₹ 15 for the first kilometre and ₹ 8 for each additional kilometre, does not form an AP as the total fare (in ₹) after each kilometre is 15, 8, 8, 8, ... . Is the statement true? Give reasons.
Answer: False. The total fare after 1 km = ₹ 15. Total fare after 2 km = \( 15 + 8 = 23 \). Total fare after 3 km = \( 23 + 8 = 31 \). The sequence of total fare is 15, 23, 31, 39, ... which forms an AP with \( a = 15 \) and \( d = 8 \). The statement saying it is 15, 8, 8, 8 is incorrect.

Question. Determine k, so that \( k^2 + 4k + 8, 2k^2 + 3k + 6 \) and \( 3k^2 + 4k + 4 \) are three consecutive terms of an AP.
Answer: For terms \( a, b, c \) to be in AP, \( 2b = a + c \).
\( 2(2k^2 + 3k + 6) = (k^2 + 4k + 8) + (3k^2 + 4k + 4) \)
\( 4k^2 + 6k + 12 = 4k^2 + 8k + 12 \)
\( 6k = 8k \Rightarrow 2k = 0 \Rightarrow k = 0 \).

Question. Show that \( (a - b)^2, (a^2 + b^2) \) and \( (a + b)^2 \) are in AP.
Answer: Let the terms be \( T_1, T_2, T_3 \).
\( T_2 - T_1 = (a^2 + b^2) - (a - b)^2 = a^2 + b^2 - (a^2 + b^2 - 2ab) = 2ab \)
\( T_3 - T_2 = (a + b)^2 - (a^2 + b^2) = (a^2 + b^2 + 2ab) - (a^2 + b^2) = 2ab \)
Since the common difference is same (\( 2ab \)), the terms are in AP.

Question. Find the 11th term from the last term (towards the first term) of the AP 12, 8, 4, ..., -84.
Answer: Reversing the AP: \( a = -84 \), \( d = 12 - 8 = 4 \) (since original \( d = -4 \), reverse \( d = +4 \)).
\( a_{11} = a + (11-1)d = -84 + 10(4) = -84 + 40 = -44 \).

Question. For the AP -3, -7, -11, ... can we find directly \( a_{30} - a_{20} \) without actually finding \( a_{30} \) and \( a_{20} \)? Give reason for your answer.
Answer: Yes. In an AP, \( a_n - a_k = (n - k)d \).
Here, \( d = -7 - (-3) = -4 \).
\( a_{30} - a_{20} = (30 - 20)d = 10(-4) = -40 \).

Question. Is 0 a term of the AP 31, 28, 25, ...? Justify your answer.
Answer: Here \( a = 31, d = 28 - 31 = -3 \).
Let \( a_n = 0 \).
\( a + (n-1)d = 0 \Rightarrow 31 + (n-1)(-3) = 0 \Rightarrow -3(n-1) = -31 \Rightarrow n-1 = \frac{31}{3} \Rightarrow n = \frac{34}{3} \).
Since \( n \) must be a natural number and \( \frac{34}{3} \) is not, 0 is not a term of this AP.

Question. If four numbers are in AP such that their sum is 50 and the greatest number is 4 times the least, then find the numbers.
Answer: Let the numbers be \( a-3d, a-d, a+d, a+3d \).
Sum = \( 4a = 50 \Rightarrow a = 12.5 \).
Greatest = 4 \(\times\) Least: \( a+3d = 4(a-3d) \Rightarrow a+3d = 4a - 12d \Rightarrow 15d = 3a \Rightarrow 5d = a \).
\( 5d = 12.5 \Rightarrow d = 2.5 \).
Numbers: \( 12.5 - 7.5, 12.5 - 2.5, 12.5 + 2.5, 12.5 + 7.5 \) which are 5, 10, 15, 20.

Question. Find the 20th term of the AP whose 7th term is 24 less than the 11th term, first term being 12.
Answer: Given \( a = 12 \) and \( a_{11} - a_7 = 24 \).
\( (a + 10d) - (a + 6d) = 24 \Rightarrow 4d = 24 \Rightarrow d = 6 \).
\( a_{20} = a + 19d = 12 + 19(6) = 12 + 114 = 126 \).

Question. If the 9th term of an AP is zero, then prove that its 29th term is twice its 19th term.
Answer: Given \( a_9 = 0 \Rightarrow a + 8d = 0 \Rightarrow a = -8d \).
\( a_{19} = a + 18d = -8d + 18d = 10d \).
\( a_{29} = a + 28d = -8d + 28d = 20d \).
Clearly, \( a_{29} = 2 \times a_{19} \).

Question. The 16th term of an AP is 1 more than twice its 8th term. If the 12th term of an AP is 47, then find its nth term.
Answer: \( a_{16} = 2a_8 + 1 \Rightarrow a+15d = 2(a+7d) + 1 \Rightarrow a+15d = 2a+14d+1 \Rightarrow d-a=1 \).
\( a_{12} = 47 \Rightarrow a+11d = 47 \).
Solving: \( a = d-1 \), so \( (d-1) + 11d = 47 \Rightarrow 12d = 48 \Rightarrow d = 4 \).
\( a = 4 - 1 = 3 \).
\( a_n = a + (n-1)d = 3 + (n-1)4 = 4n - 1 \).

Question. Find the 19th term of the following sequence. \( t_n = \begin{cases} n^2, & \text{where } n \text{ is even} \\ n^2 - 1, & \text{where } n \text{ is odd} \end{cases} \)
Answer: Since \( n = 19 \) is odd, we use the formula \( t_n = n^2 - 1 \).
\( t_{19} = 19^2 - 1 = 361 - 1 = 360 \).

Question. Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.
Answer: Let parts be \( a-d, a, a+d \).
Sum = \( 3a = 207 \Rightarrow a = 69 \).
Smaller parts are \( a-d \) and \( a \).
\( (a-d)a = 4623 \Rightarrow (69-d)69 = 4623 \Rightarrow 69-d = 67 \Rightarrow d = 2 \).
Parts are \( 69-2, 69, 69+2 \), i.e., 67, 69, 71.