CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set 14

Question. If \( a, b, c, d, e \) and \( f \) are in AP with common difference 3, then the value of \( e - c \) is
(a) 3
(b) 5
(c) 6
(d) 9
Answer: (c) 6

Question. The 10th term of the sequence \( \sqrt{3}, \sqrt{12}, \sqrt{27}, \dots \) is
(a) \( 10\sqrt{2} \)
(b) \( 10\sqrt{3} \)
(c) \( 10\sqrt{5} \)
(d) \( 10\sqrt{7} \)
Answer: (b) \( 10\sqrt{3} \)

Question. If second term of an AP is \( (x - y) \) and 5th term is \( (x + y) \), then its first term is
(a) \( -\frac{1}{3}y \)
(b) \( x - \frac{2}{3}y \)
(c) \( x - \frac{4}{3}y \)
(d) \( x - \frac{5}{3}y \)
Answer: (d) \( x - \frac{5}{3}y \)

Question. A man receives Rs. 60 for the first week and Rs. 3 more each week than the preceding week. Then his earning by the 20th week is
(a) Rs. 1767
(b) Rs. 1770
(c) Rs. 1780
(d) Rs. 1787
Answer: (b) Rs. 1770

Question. The first positive term of the AP \( -11, -8, -5, \dots \) is
(a) 5th term
(b) 4th term
(c) 3rd term
(d) 6th term
Answer: (a) 5th term

Question. Which of the following is not an AP?
(a) \( 1, 4, 7, 10, \dots \)
(b) \( -5, -2, 1, 4, \dots \)
(c) \( 3, 7, 12, 18, \dots \)
(d) \( 11, 14, 17, 20, \dots \)
Answer: (c) \( 3, 7, 12, 18, \dots \)

Question. If the 17th term of an AP exceeds the 13th term of the AP by 15, then the common difference is
(a) 3
(b) 3.75
(c) 3.5
(d) 4.25
Answer: (b) 3.75

Question. If for an AP, \( a_5 = a_{10} = 71 \), then \( a_{15} \) is
(a) 72
(b) 71
(c) 76
(d) 66
Answer: (b) 71

Question. The sum of first 20 natural numbers is
(a) 110
(b) 170
(c) 190
(d) 210
Answer: (d) 210

Question. If \( a_k \) be the kth term of the AP \( 3, 15, 27, 39, \dots \) and value of \( a_k \) is 180 more than the value of \( a_{50} \), then \( k = \)
(a) 58
(b) 62
(c) 65
(d) 68
Answer: (c) 65

Question. The sum of \( 2 + 5 + 8 + \dots + 152 \) is
(a) 3924
(b) 3927
(c) 3936
(d) 3942
Answer: (b) 3927

Question. If sum to \( n \) terms of an AP is \( 3n^2 + 4n \), then the common difference of the AP is
(a) 3
(b) 4
(c) 6
(d) 7
Answer: (c) 6

Question. If \( S_n \) denotes the sum of first \( n \) terms of an AP, whose common difference is \( d \), then \( S_n - 2S_{n-1} + S_{n-2} \) \( (n \geq 3) \) is equal to
(a) \( t_n \)
(b) \( t_{n-1} \)
(c) \( d \)
(d) None of the options
Answer: (c) \( d \)

Question. Find the number of three-digit natural numbers which are divisible by 11. [Foreign 2013]
Answer: Three-digit numbers divisible by 11 are \( 110, 121, 132, \dots, 990 \).
Here, \( a = 110 \), \( d = 11 \), and \( a_n = 990 \).
\( a_n = a + (n - 1)d \)
\( \implies 990 = 110 + (n - 1)11 \)
\( \implies 880 = (n - 1)11 \)
\( \implies n - 1 = 80 \)
\( \implies n = 81 \)
Thus, there are 81 such numbers.

Question. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5. [AI 2014]
Answer: Numbers divisible by both 2 and 5 must be divisible by 10.
The sequence of such numbers between 101 and 999 is \( 110, 120, \dots, 990 \).
Here, \( a = 110 \), \( d = 10 \), and \( a_n = 990 \).
\( a_n = a + (n - 1)d \)
\( \implies 990 = 110 + (n - 1)10 \)
\( \implies 880 = (n - 1)10 \)
\( \implies n - 1 = 88 \)
\( \implies n = 89 \)
Thus, there are 89 such numbers.

Question. In an AP, if \( S_5 + S_7 = 167 \) and \( S_{10} = 235 \), then find the AP, where \( S_n \) denotes the sum of its first \( n \) terms. [AI 2015]
Answer: Let the first term be \( a \) and common difference be \( d \).
\( S_{10} = \frac{10}{2}[2a + 9d] = 235 \implies 5(2a + 9d) = 235 \implies 2a + 9d = 47 \) --- (i)
\( S_5 + S_7 = \frac{5}{2}[2a + 4d] + \frac{7}{2}[2a + 6d] = 167 \)
\( \implies 5(a + 2d) + 7(a + 3d) = 167 \)
\( \implies 5a + 10d + 7a + 21d = 167 \implies 12a + 31d = 167 \) --- (ii)
Multiplying (i) by 6, we get \( 12a + 54d = 282 \).
Subtracting (ii) from this, \( 23d = 115 \implies d = 5 \).
Substituting \( d = 5 \) in (i): \( 2a + 45 = 47 \implies 2a = 2 \implies a = 1 \).
The AP is \( 1, 6, 11, \dots \)

Question. If the ratio of sum of the first \( m \) and \( n \) terms of an A.P. is \( m^2 : n^2 \), show that the ratio of its \( m \)th and \( n \)th terms is \( (2m - 1) : (2n - 1) \). [Foreign 2016]
Answer: Given \( \frac{S_m}{S_n} = \frac{m^2}{n^2} \)
\( \implies \frac{\frac{m}{2}[2a + (m-1)d]}{\frac{n}{2}[2a + (n-1)d]} = \frac{m^2}{n^2} \)
\( \implies \frac{2a + (m-1)d}{2a + (n-1)d} = \frac{m}{n} \)
Replace \( m \) by \( 2m - 1 \) and \( n \) by \( 2n - 1 \):
\( \implies \frac{2a + (2m-2)d}{2a + (2n-2)d} = \frac{2m-1}{2n-1} \)
\( \implies \frac{a + (m-1)d}{a + (n-1)d} = \frac{2m-1}{2n-1} \)
\( \implies \frac{a_m}{a_n} = \frac{2m-1}{2n-1} \). Hence Proved.

Question. The first term of an AP is 5 and its 100th term is \( -292 \). Find the 50th term.
Answer: \( a = 5 \), \( a_{100} = -292 \).
\( a_{100} = a + 99d = -292 \)
\( \implies 5 + 99d = -292 \implies 99d = -297 \implies d = -3 \).
\( a_{50} = a + 49d = 5 + 49(-3) = 5 - 147 = -142 \).

Question. Animation is a method in which a sequence of images are manipulated to appear as moving objects. An animation specialist wants to show the growth of a sapling into a tree through animation. She follows the steps below:

• She develops the first image by designing a sapling containing a certain number of leaves.
• She develops the second image by adding 15 leaves to the first image.
• She develops the third image by adding 22 leaves to the second image.
• Then the fourth image by adding 29 leaves to the third image and so on.

If she continues the process in the same manner, how many leaves will she be adding to the 25th image to develop the 26th image? Show your work. [CFPQ, CBSE]
Answer: The number of leaves added at each step forms an AP: \( 15, 22, 29, \dots \)
First term \( a = 15 \), Common difference \( d = 22 - 15 = 7 \).
To develop the 26th image from the 25th image, she is adding the 25th value in this addition sequence.
Number of leaves added \( = a_{25} = a + 24d \)
\( \implies a_{25} = 15 + 24(7) = 15 + 168 = 183 \).
She will be adding 183 leaves.

Question. Find \( a \) and \( b \) so that the numbers \( a, 7, b, 23 \) are in A.P. [CBSE 2022]
Answer: Since \( a, 7, b, 23 \) are in AP:
\( a_2 = 7 \) and \( a_4 = 23 \).
\( a_4 - a_2 = 2d \implies 23 - 7 = 2d \implies 16 = 2d \implies d = 8 \).
Now, \( b = a_2 + d = 7 + 8 = 15 \).
And \( a = a_2 - d = 7 - 8 = -1 \).
So, \( a = -1 \) and \( b = 15 \).

Question. Find the sum of all multiples of 7 lying between 500 and 900. [AI 2012]
Answer: Multiples of 7 between 500 and 900 are \( 504, 511, \dots, 896 \).
Here \( a = 504 \), \( d = 7 \), \( a_n = 896 \).
\( 896 = 504 + (n-1)7 \)
\( \implies 392 = (n-1)7 \implies n-1 = 56 \implies n = 57 \).
Sum \( S_n = \frac{n}{2}(a + a_n) = \frac{57}{2}(504 + 896) = \frac{57 \times 1400}{2} = 57 \times 700 = 39900 \).

Question. If the seventh term of an AP is \( \frac{1}{9} \) and its ninth term is \( \frac{1}{7} \), find its 63rd term. [Delhi 2014]
Answer: \( a_7 = a + 6d = \frac{1}{9} \) --- (i)
\( a_9 = a + 8d = \frac{1}{7} \) --- (ii)
Subtracting (i) from (ii): \( 2d = \frac{1}{7} - \frac{1}{9} = \frac{2}{63} \implies d = \frac{1}{63} \).
Substituting \( d \) in (i): \( a + 6(\frac{1}{63}) = \frac{1}{9} \implies a = \frac{7}{63} - \frac{6}{63} = \frac{1}{63} \).
\( a_{63} = a + 62d = \frac{1}{63} + 62(\frac{1}{63}) = \frac{63}{63} = 1 \).

Question. The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P. [Foreign 2014, DoE]
Answer: \( S_7 = 63 \implies \frac{7}{2}[2a + 6d] = 63 \implies a + 3d = 9 \) --- (i)
Sum of next 7 terms is 161, so \( S_{14} = 63 + 161 = 224 \).
\( S_{14} = \frac{14}{2}[2a + 13d] = 224 \implies 7[2a + 13d] = 224 \implies 2a + 13d = 32 \) --- (ii)
Multiply (i) by 2: \( 2a + 6d = 18 \).
Subtract from (ii): \( 7d = 14 \implies d = 2 \).
From (i), \( a + 3(2) = 9 \implies a = 3 \).
\( a_{28} = a + 27d = 3 + 27(2) = 3 + 54 = 57 \).

Question. In an AP, if the 12th term is \( -13 \) and the sum of its first four terms is 24, find the sum of its first ten terms. [Foreign 2015]
Answer: \( a_{12} = a + 11d = -13 \) --- (i)
\( S_4 = \frac{4}{2}[2a + 3d] = 24 \implies 2a + 3d = 12 \) --- (ii)
From (i), \( a = -13 - 11d \). Substitute in (ii):
\( 2(-13 - 11d) + 3d = 12 \implies -26 - 22d + 3d = 12 \)
\( \implies -19d = 38 \implies d = -2 \).
Then \( a = -13 - 11(-2) = 9 \).
\( S_{10} = \frac{10}{2}[2(9) + 9(-2)] = 5[18 - 18] = 0 \).

Question. Divide 56 in four parts in AP, such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is \( 5 : 6 \). [Foreign 2016]
Answer: Let the parts be \( a - 3d, a - d, a + d, a + 3d \).
Sum \( = 4a = 56 \implies a = 14 \).
Given: \( \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{5}{6} \implies \frac{a^2 - 9d^2}{a^2 - d^2} = \frac{5}{6} \)
\( \implies 6a^2 - 54d^2 = 5a^2 - 5d^2 \implies a^2 = 49d^2 \)
\( \implies 14^2 = 49d^2 \implies 196 = 49d^2 \implies d^2 = 4 \implies d = \pm 2 \).
The parts are \( 14 - 6, 14 - 2, 14 + 2, 14 + 6 \) i.e., \( 8, 12, 16, 20 \).

Question. The digits of a positive number of three digits are in AP and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number. [AI 2016]
Answer: Let the digits be \( a - d, a, a + d \).
Sum \( = 3a = 15 \implies a = 5 \).
Original number \( = 100(a - d) + 10a + (a + d) = 111a - 99d \).
Reversed number \( = 100(a + d) + 10a + (a - d) = 111a + 99d \).
Given: Original \( - \) Reversed \( = 594 \)
\( \implies (111a - 99d) - (111a + 99d) = 594 \)
\( \implies -198d = 594 \implies d = -3 \).
Digits are \( 5 - (-3), 5, 5 + (-3) \implies 8, 5, 2 \).
The number is 852.

Question. Given below are the details of an experiment using a bucket and a mug to understand water consumption. The bucket’s volume is 30 litres and the mug’s volume is \( \frac{1}{20} \) of the bucket. The bucket has 1 litre of water before the tap is turned on. The tap is filling the bucket at a constant rate of 0.1 litres per second. Every 30 seconds, he takes a mug full of water from the bucket.
(i) Write an arithmetic progression for the volume of water in the bucket every 30 seconds.
(ii) Find the volume of water in the bucket after exactly 7.5 minutes. Show your work.
(Note: Assume no spillage of water) [CFPQ, CBSE]
Answer: (i) Mug volume \( = \frac{1}{20} \times 30 = 1.5 \) L.
In 30 seconds, water filled by tap \( = 0.1 \times 30 = 3 \) L.
Net water change every 30 seconds \( = 3 - 1.5 = 1.5 \) L.
Initial volume \( = 1 \) L.
AP for volume every 30 seconds: \( 1, 2.5, 4, 5.5, \dots \)
(ii) \( 7.5 \) minutes \( = 7.5 \times 60 = 450 \) seconds.
Number of 30-sec intervals \( = \frac{450}{30} = 15 \).
Volume after 15 intervals \( = a + 15d = 1 + 15(1.5) = 1 + 22.5 = 23.5 \) L.

Question. The 14th term of an AP is twice its 8th term. If the 6th term is \( -8 \), then find the sum of its first 20 terms.
Answer: \( a + 13d = 2(a + 7d) \implies a + 13d = 2a + 14d \implies a = -d \) --- (i)
\( a_6 = a + 5d = -8 \) --- (ii)
Substitute (i) in (ii): \( -d + 5d = -8 \implies 4d = -8 \implies d = -2 \).
Then \( a = 2 \).
\( S_{20} = \frac{20}{2}[2(2) + 19(-2)] = 10[4 - 38] = 10(-34) = -340 \).

Question. If \( T_1, T_2, T_3, \dots, T_n \) are consecutive terms of an AP, then prove that \( \frac{1}{T_1 T_2} + \frac{1}{T_2 T_3} + \dots + \frac{1}{T_{n-1} T_n} = \frac{n - 1}{T_1 T_n} \). [Hots]
Answer: Let \( d \) be the common difference.
LHS \( = \frac{1}{d} [ \frac{d}{T_1 T_2} + \frac{d}{T_2 T_3} + \dots + \frac{d}{T_{n-1} T_n} ] \)
\( \implies \frac{1}{d} [ \frac{T_2 - T_1}{T_1 T_2} + \frac{T_3 - T_2}{T_2 T_3} + \dots + \frac{T_n - T_{n-1}}{T_{n-1} T_n} ] \)
\( \implies \frac{1}{d} [ (\frac{1}{T_1} - \frac{1}{T_2}) + (\frac{1}{T_2} - \frac{1}{T_3}) + \dots + (\frac{1}{T_{n-1}} - \frac{1}{T_n}) ] \)
\( \implies \frac{1}{d} [ \frac{1}{T_1} - \frac{1}{T_n} ] = \frac{1}{d} [ \frac{T_n - T_1}{T_1 T_n} ] \)
\( \implies \frac{1}{d} [ \frac{(n-1)d}{T_1 T_n} ] = \frac{n-1}{T_1 T_n} = \) RHS. Hence Proved.

Question. Along a road lies an odd number of stones placed at intervals of 10 metres. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried the job with one of the end stones by carrying them in succession. In carrying all the stones he covered a distance of 3 km. Find the number of stones. [Hots]
Answer: Let there be \( 2n + 1 \) stones. Middle stone is at pos 0. Stones on one side are at \( 10, 20, \dots, 10n \) and on other side at \( -10, -20, \dots, -10n \).
Man starts at end stone \( -10n \).
Dist to bring end stone to middle \( = 10n \).
Then for each of other \( 2n \) stones, he goes from middle to stone and back.
Total dist \( = 10n + [2 \times 10(n-1) + \dots + 2 \times 10(1)] + [2 \times 10(1) + \dots + 2 \times 10(n)] = 3000 \)
\( \implies 10n + 20 \frac{(n-1)n}{2} + 20 \frac{n(n+1)}{2} = 3000 \)
\( \implies 10n + 10n^2 - 10n + 10n^2 + 10n = 3000 \)
\( \implies 20n^2 + 10n - 3000 = 0 \implies 2n^2 + n - 300 = 0 \).
Solving gives \( n = 12 \).
Total stones \( = 2n + 1 = 2(12) + 1 = 25 \).

Question. If \( S_n \) denotes the sum of the first \( n \) terms of an AP, prove that \( S_{30} = 3(S_{20} - S_{10}) \). [Foreign 2014]
Answer: RHS \( = 3 [ \frac{20}{2}(2a + 19d) - \frac{10}{2}(2a + 9d) ] \)
\( \implies 3 [ 20a + 190d - 10a - 45d ] = 3 [ 10a + 145d ] = 30a + 435d \)
LHS \( = S_{30} = \frac{30}{2}(2a + 29d) = 15(2a + 29d) = 30a + 435d \).
LHS \( = \) RHS. Hence Proved.

Question. Find the 60th term of the AP \( 8, 10, 12, \dots \), if it has a total of 60 terms and hence find the sum of its last 10 terms. [AI 2015]
Answer: \( a = 8 \), \( d = 2 \).
\( a_{60} = 8 + 59(2) = 126 \).
To find sum of last 10 terms (terms 51 to 60):
\( a_{51} = 8 + 50(2) = 108 \).
Sum \( = \frac{10}{2}(108 + 126) = 5(234) = 1170 \).

Question. Find the middle term of the sequence formed by all three-digit numbers which leave a remainder 3, when divided by 4. Also, find the sum of all numbers on both sides of the middle terms separately. [Foreign 2015]
Answer: Sequence: \( 103, 107, \dots, 999 \).
\( 999 = 103 + (n-1)4 \implies n = 225 \).
Middle term is \( \frac{225+1}{2} = 113 \)th term.
\( a_{113} = 103 + 112(4) = 551 \).
Sum before middle term (1 to 112): \( \frac{112}{2}[2(103) + 111(4)] = 56[206 + 444] = 36400 \).
Sum after middle term (114 to 225): \( \frac{112}{2}[2(555) + 111(4)] = 56[1110 + 444] = 87024 \).

Question. A thief runs with a uniform speed of 100 m/minute. After one minute, a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the police man will catch the thief? [Delhi 2016]
Answer: Let the policeman catch the thief in \( n \) minutes. The thief has been running for \( (n + 1) \) minutes.
Distance by thief \( = 100(n + 1) \)
Distance by policeman \( = \frac{n}{2}[2(100) + (n - 1)10] = 100n + 5n(n - 1) = 5n^2 + 95n \)
\( \implies 5n^2 + 95n = 100n + 100 \implies 5n^2 - 5n - 100 = 0 \implies n^2 - n - 20 = 0 \)
\( \implies (n - 5)(n + 4) = 0 \implies n = 5 \).
Policeman will catch the thief in 5 minutes.

Question. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of \( X \) such that sum of numbers of houses preceding the house numbered \( X \) is equal to sum of the numbers of houses following \( X \). [AI 2016]
Answer: Given \( S_{X-1} = S_{49} - S_X \)
\( \implies \frac{X-1}{2}(1 + X-1) = \frac{49}{2}(1 + 49) - \frac{X}{2}(1 + X) \)
\( \implies \frac{X(X-1)}{2} = 1225 - \frac{X^2 + X}{2} \)
\( \implies X^2 - X = 2450 - X^2 - X \implies 2X^2 = 2450 \implies X^2 = 1225 \)
\( \implies X = 35 \).
Value of \( X \) is 35.

Question. Reshma wanted to save at least Rs. 6,500 for sending her daughter to school next year (after 12 months). She saved Rs. 450 in the first month and raised her savings by Rs. 20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year? [Foreign 2016]
Answer: \( a = 450 \), \( d = 20 \), \( n = 12 \).
\( S_{12} = \frac{12}{2}[2(450) + 11(20)] = 6[900 + 220] = 6[1120] = 6720 \).
She will save Rs. 6720. Since \( 6720 > 6500 \), she will be able to send her daughter to school.

Assertion and reason questions

Question. Assertion (A): 10th term of an AP is 41 whose first term is 5 and common difference is 4.
Reason (R): nth term of an AP is \( a_n = a + (n - 1)d \), where \( a = \) first term, \( d = \) common difference.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation of A.

Question. Assertion (A): In an AP with \( a = 15 \), \( d = -3 \) then 6th term will be zero.
Reason (R): \( a - d, a, a + d \) are three numbers in AP.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (b) Both A and R are true but R is not the correct explanation of A.