CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set 15

CASE-BASED QUESTIONS

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.

Based on the above information, answer the following questions:

Question. Find the production during first year.
Answer: Let the production in the first year be \( a \) and the annual increase be \( d \).
Given: \( a_6 = 16000 \) and \( a_9 = 22600 \).
\( a + 5d = 16000 \) --- (i)
\( a + 8d = 22600 \) --- (ii)
Subtracting (i) from (ii):
\( 3d = 6600 \)
\( \implies d = 2200 \)
Substitute \( d \) in (i):
\( a + 5(2200) = 16000 \)
\( \implies a + 11000 = 16000 \)
\( \implies a = 5000 \)
The production during the first year was 5000 sets.

Question. Find the production during 8th year.
Answer: Production during 8th year \( = a_8 = a + 7d \)
\( a_8 = 5000 + 7(2200) \)
\( \implies a_8 = 5000 + 15400 = 20400 \).
The production during the 8th year was 20,400 sets.

Question. In which year, the production is 29,200.
OR
Find the difference of the production during 7th year and 4th year.
Answer: Let the year be \( n \).
\( a_n = a + (n - 1)d = 29200 \)
\( \implies 5000 + (n - 1)2200 = 29200 \)
\( \implies (n - 1)2200 = 24200 \)
\( \implies n - 1 = 11 \)
\( \implies n = 12 \)
The production will be 29,200 in the 12th year.
OR
Difference \( = a_7 - a_4 = (a + 6d) - (a + 3d) = 3d \)
\( \text{Difference} = 3 \times 2200 = 6600 \).
The difference of the production is 6,600 sets.

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. What is the minimum number of days he needs to practice till his goal is achieved?
Answer: Here, first term \( a = 51 \), common difference \( d = -2 \), and last term \( a_n = 31 \).
\( a_n = a + (n - 1)d \)
\( \implies 31 = 51 + (n - 1)(-2) \)
\( \implies -20 = -2(n - 1) \)
\( \implies n - 1 = 10 \)
\( \implies n = 11 \)
He needs to practice for 11 days.

Question. If nth term of an AP is given by \( a_n = 2n + 3 \), then find common difference of an AP.
Answer: \( a_n = 2n + 3 \)
For \( n = 1 \), \( a_1 = 2(1) + 3 = 5 \)
For \( n = 2 \), \( a_2 = 2(2) + 3 = 7 \)
Common difference \( d = a_2 - a_1 = 7 - 5 = 2 \).

Question. Find the time taken by veer to run 200 m on the 6th day of practice.
OR
Find the value of \( x \), for which \( 2x, x + 10, 3x + 2 \) are three consecutive terms of an AP.
Answer: Time on 6th day \( = a_6 = a + 5d \)
\( a_6 = 51 + 5(-2) = 51 - 10 = 41 \) seconds.
OR
For consecutive terms: \( 2(x + 10) = 2x + (3x + 2) \)
\( \implies 2x + 20 = 5x + 2 \)
\( \implies 3x = 18 \)
\( \implies x = 6 \).

Rampal deposits some money in bank and gets an increment on it every year. For example, Rampal deposits some money Rs. \( p \) in bank and get an interest of Rs. \( I \) on it every year then this is represented as \( p, p + I, p + 2I, p + 3I \dots \) The sequence \( p, p + I, p + 2I, p + 3I \dots \) forms an AP, with 1st term \( p \) and common difference \( I \). Using the AP apply formula and its application. Answer the questions based on above:

Question. If \( 4p + 8, 2p^2 + 3p + 6 \) and \( 3p^2 + 4p + 4 \) form three consecutive terms of an AP, then find \( p \).
Answer: For terms in AP: \( 2(2p^2 + 3p + 6) = (4p + 8) + (3p^2 + 4p + 4) \)
\( \implies 4p^2 + 6p + 12 = 3p^2 + 8p + 12 \)
\( \implies p^2 - 2p = 0 \)
\( \implies p(p - 2) = 0 \)
Since money deposit \( p \neq 0 \), therefore \( p = 2 \).

Question. Find the 12th term of AP: \( 10.0, 10.5, 11.0, 11.5, \dots \)
Answer: Here, \( a = 10.0 \), \( d = 10.5 - 10.0 = 0.5 \).
\( a_{12} = a + 11d = 10.0 + 11(0.5) \)
\( \implies a_{12} = 10.0 + 5.5 = 15.5 \).

Question. The 17th term of an AP exceeds its 10th term by 14, then find the common difference.
OR
Find the qth term of the AP: \( \frac{1}{p}, \frac{1 + p}{p}, \frac{1 + 2p}{p}, \dots \)
Answer: Given: \( a_{17} = a_{10} + 14 \)
\( \implies (a + 16d) = (a + 9d) + 14 \)
\( \implies 7d = 14 \)
\( \implies d = 2 \).
The common difference is 2.
OR
First term \( a = \frac{1}{p} \), Common difference \( d = \frac{1 + p}{p} - \frac{1}{p} = \frac{p}{p} = 1 \).
\( a_q = a + (q - 1)d = \frac{1}{p} + (q - 1)(1) \)
\( \implies a_q = \frac{1 + p(q - 1)}{p} = \frac{pq - p + 1}{p} \).

Assess Yourself

Question. The 10th term of the sequence \( \sqrt{3}, \sqrt{12}, \sqrt{27}, \dots \) is
(a) \( \sqrt{243} \)
(b) \( \sqrt{300} \)
(c) \( \sqrt{363} \)
(d) \( \sqrt{432} \)
Answer: (b) \( \sqrt{300} \)

Question. If \( x, y, z \) are in AP, then the value of \( (x + y - z)(y + z - x) \) is
(a) \( 8yz - 3y^2 - 4z^2 \)
(b) \( 4xz + 3y^2 \)
(c) \( 8xy + 4x^2 - 3y^2 \)
(d) None of the options
Answer: (a) \( 8yz - 3y^2 - 4z^2 \)

Question. The second term of an AP is \( (x - y) \) and 5th term is \( (x + y) \), its first term is
(a) \( x - \frac{1}{3}y \)
(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. The 15th term of the sequence \( x - 7, x - 2, x + 3, \dots \) is
(a) \( x + 63 \)
(b) \( x + 73 \)
(c) \( x + 83 \)
(d) \( x + 53 \)
Answer: (a) \( x + 63 \)

Question. The sum of \( n \) terms of the series \( 2, 5, 8, 11, \dots \) is 60100, then \( n \) is
(a) 100
(b) 200
(c) 150
(d) 250
Answer: (b) 200

Question. If \( S_n = nP + \frac{n(n - 1)}{2}Q \), where \( S_n \) denotes the sum of the first \( n \) terms of an AP, then common difference is
(a) \( P + Q \)
(b) \( 2P + 3Q \)
(c) \( 2Q \)
(d) \( Q \)
Answer: (d) \( Q \)

Question. If \( a_1, a_2, a_3, \dots \) is an AP such that \( a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 300 \), then \( a_1 + a_2 + a_3 + \dots + a_{24} \) is equal to
(a) 1200
(b) 400
(c) 800
(d) 1000
Answer: (a) 1200

Question. Determine \( k \) so that \( k + 2, 4k - 6 \) and \( 3k - 2 \) are the three consecutive terms of an AP.
Answer: \( 2(4k - 6) = (k + 2) + (3k - 2) \)
\( \implies 8k - 12 = 4k \)
\( \implies 4k = 12 \)
\( \implies k = 3 \).

Question. The sum of the 5th and 7th terms of an AP is 52 and the 10th term is 46. Find the AP.
Answer: \( a_5 + a_7 = 52 \implies 2a + 10d = 52 \implies a + 5d = 26 \)
\( a_{10} = 46 \implies a + 9d = 46 \)
Subtracting: \( 4d = 20 \implies d = 5 \).
\( a + 25 = 26 \implies a = 1 \).
The AP is \( 1, 6, 11, \dots \)

Question. If the sum of first \( n \) terms of an AP is given by \( S_n = 4n^2 - 3n \), find the nth term of the AP.
Answer: \( a_n = S_n - S_{n-1} \)
\( a_n = (4n^2 - 3n) - [4(n - 1)^2 - 3(n - 1)] \)
\( a_n = 4n^2 - 3n - [4(n^2 - 2n + 1) - 3n + 3] \)
\( a_n = 4n^2 - 3n - [4n^2 - 8n + 4 - 3n + 3] \)
\( a_n = 4n^2 - 3n - 4n^2 + 11n - 7 \)
\( a_n = 8n - 7 \).

Question. If 5th term of an AP is zero, show that its 33rd term is four times its 12th term.
Answer: Given \( a_5 = a + 4d = 0 \implies a = -4d \).
\( a_{12} = a + 11d = -4d + 11d = 7d \)
\( a_{33} = a + 32d = -4d + 32d = 28d \)
\( a_{33} = 4 \times 7d = 4 \times a_{12} \). Hence Proved.

Question. Find 10th term from the last of the AP \( 8, 10, 12, \dots, 126 \).
Answer: Last term \( L = 126 \), common difference \( d = 2 \).
Term from last \( = L - (n - 1)d \)
\( \text{10th term from last} = 126 - (10 - 1)2 = 126 - 18 = 108 \).

Question. The sum of 5th and 9th terms of an AP is 72 and the sum of 7th and 12th terms is 97. Find the AP.
Answer: \( a_5 + a_9 = 72 \implies 2a + 12d = 72 \implies a + 6d = 36 \) --- (i)
\( a_7 + a_{12} = 97 \implies 2a + 17d = 97 \) --- (ii)
Multiply (i) by 2: \( 2a + 12d = 72 \)
Subtract from (ii): \( 5d = 25 \implies d = 5 \).
From (i), \( a + 30 = 36 \implies a = 6 \).
The AP is \( 6, 11, 16, \dots \)

Question. If \( T_m, T_{m + n} \) and \( T_{m - n} \) are respectively the \( m \)th, \( (m + n) \)th and \( (m - n) \)th term of an AP, then prove that \( T_{m + n} + T_{m - n} = 2 T_m \).  
Answer: \( T_{m + n} = a + (m + n - 1)d \)
\( T_{m - n} = a + (m - n - 1)d \)
Adding both: \( T_{m + n} + T_{m - n} = 2a + (m + n - 1 + m - n - 1)d \)
\( \implies T_{m + n} + T_{m - n} = 2a + (2m - 2)d = 2[a + (m - 1)d] = 2 T_m \). Hence Proved.

Question. If the sum of first \( m \) terms of an AP is \( n \) and the sum of first \( n \) terms is \( m \), then show that the sum of first \( (m + n) \) terms is \( -(m + n) \).  
Answer: Given \( S_m = n \) and \( S_n = m \).
\( \frac{m}{2}[2a + (m - 1)d] = n \implies 2a + (m - 1)d = \frac{2n}{m} \) --- (i)
\( \frac{n}{2}[2a + (n - 1)d] = m \implies 2a + (n - 1)d = \frac{2m}{n} \) --- (ii)
Subtracting (ii) from (i): \( (m - n)d = \frac{2n}{m} - \frac{2m}{n} = \frac{2(n^2 - m^2)}{mn} = \frac{-2(m - n)(m + n)}{mn} \)
\( \implies d = \frac{-2(m + n)}{mn} \).
Substituting \( d \) and finding \( 2a \), then applying in \( S_{m+n} = \frac{m+n}{2}[2a + (m+n-1)d] \), we get \( S_{m+n} = -(m+n) \). Hence Proved.

Question. The angles of a quadrilateral are in AP. The greatest angle is double the least. Find all the four angles. 
Answer: Let the angles be \( a - 3d, a - d, a + d, a + 3d \).
Sum \( = 4a = 360^\circ \implies a = 90^\circ \).
Given: \( a + 3d = 2(a - 3d) \)
\( \implies 90 + 3d = 180 - 6d \)
\( \implies 9d = 90 \implies d = 10^\circ \).
The angles are \( 60^\circ, 80^\circ, 100^\circ, 120^\circ \).

Question. Find the sum of \( 2n \) terms of the series \( 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + \dots \).  
Answer: Series \( = (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + \dots \) to \( n \) pairs.
\( = -1(3) - 1(7) - 1(11) - \dots \)
\( = -(3 + 7 + 11 + \dots ) \) to \( n \) terms.
Sum \( = -[\frac{n}{2}\{2(3) + (n - 1)4\}] = -[\frac{n}{2}\{6 + 4n - 4\}] = -[\frac{n}{2}(2 + 4n)] = -n(2n + 1) \).

Question. If sum of first \( k \) terms of an AP is \( 2k^2 + 3k \), then what is its second term?
Answer: \( S_k = 2k^2 + 3k \)
\( a_1 = S_1 = 2(1)^2 + 3(1) = 5 \)
\( S_2 = 2(2)^2 + 3(2) = 8 + 6 = 14 \)
\( a_2 = S_2 - S_1 = 14 - 5 = 9 \).
The second term is 9.

Question. Check whether 301 is a term of the list of numbers 5, 11, 17, 23, \dots .
Answer: Here \( a = 5, d = 6 \).
Let \( a_n = 301 \)
\( \implies 5 + (n - 1)6 = 301 \)
\( \implies 6(n - 1) = 296 \)
\( \implies n - 1 = \frac{296}{6} = 49.33 \)
Since \( n \) is not an integer, 301 is not a term.

Question. If sum of all the terms of an AP 1, 4, 7, 10, \dots, \( x \) is 287, find \( x \).
Answer: \( a = 1, d = 3, S_n = 287 \).
\( \frac{n}{2}[2(1) + (n - 1)3] = 287 \implies n(3n - 1) = 574 \)
\( \implies 3n^2 - n - 574 = 0 \)
Solving quadratic: \( n = 14 \).
\( x = a_{14} = 1 + 13(3) = 1 + 39 = 40 \).

Question. The sum of four numbers in AP is 26 and the sum of their squares is 214. Find the numbers.
Answer: Let the numbers be \( a - 3d, a - d, a + d, a + 3d \).
Sum \( = 4a = 26 \implies a = 6.5 \).
Sum of squares \( = (a - 3d)^2 + (a - d)^2 + (a + d)^2 + (a + 3d)^2 = 214 \)
\( \implies 4a^2 + 20d^2 = 214 \)
\( \implies 4(6.5)^2 + 20d^2 = 214 \implies 169 + 20d^2 = 214 \)
\( \implies 20d^2 = 45 \implies d^2 = 2.25 \implies d = 1.5 \).
Numbers are \( 2, 5, 8, 11 \).

Question. The sums of \( n \) terms of two APs are in the ratio \( 5n + 4 : 9n + 6 \). Find the ratio of their 25th terms.
Answer: \( \frac{S_n}{S'_n} = \frac{5n + 4}{9n + 6} \)
Ratio of 25th terms \( = \frac{a_{25}}{a'_{25}} = \text{replace } n \text{ by } 2(25) - 1 = 49 \)
\( \text{Ratio} = \frac{5(49) + 4}{9(49) + 6} = \frac{245 + 4}{441 + 6} = \frac{249}{447} = \frac{83}{149} \).

Question. Find sum of all three digit numbers which leave the remainder 3 when divided by 5.
Answer: Numbers are \( 103, 108, 113, \dots, 998 \).
\( a = 103, d = 5, a_n = 998 \)
\( 998 = 103 + (n - 1)5 \implies n - 1 = 179 \implies n = 180 \).
\( \text{Sum} = \frac{180}{2}(103 + 998) = 90 \times 1101 = 99090 \).

Question. A polygon has 31 sides, the lengths of which, starting from the smallest, are in AP. If the perimeter of the polygon is 527 cm and the length of the largest side is sixteen times the smallest, find the length of the smallest side and the common difference of the AP.
Answer: \( n = 31, S_{31} = 527, a_{31} = 16a \).
\( S_{31} = \frac{31}{2}(a + a_{31}) = 527 \)
\( \implies \frac{31}{2}(a + 16a) = 527 \implies \frac{31 \times 17a}{2} = 527 \)
\( \implies 527a = 1054 \implies a = 2 \text{ cm} \).
\( a_{31} = a + 30d \implies 32 = 2 + 30d \implies 30d = 30 \implies d = 1 \text{ cm} \).
Smallest side is 2 cm and common difference is 1 cm.

Question. Renu Suman donated Rs. 3,50,000 to a school from her life long savings for giving 7 cash prizes to students for their academic performances. If the worth of each prize is Rs. 10,000 less than the worth of its preceding prize, find the worth of each prize.
Answer: \( n = 7, S_7 = 350000, d = -10000 \).
\( 350000 = \frac{7}{2}[2a + 6(-10000)] \implies 50000 = a - 30000 \implies a = 80000 \).
The prizes are Rs. 80,000, Rs. 70,000, Rs. 60,000, Rs. 50,000, Rs. 40,000, Rs. 30,000, and Rs. 20,000.

Question. The sum of three numbers in an AP is \( -3 \), and their product is 8. Find the numbers.
Answer: Let numbers be \( a - d, a, a + d \).
Sum \( = 3a = -3 \implies a = -1 \).
Product \( = (a - d)a(a + d) = a(a^2 - d^2) = 8 \)
\( \implies -1(1 - d^2) = 8 \implies d^2 - 1 = 8 \implies d^2 = 9 \implies d = \pm 3 \).
The numbers are \( -4, -1, 2 \).