CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set G

OBJECTIVE TYPE QUESTIONS

Question. The equation \( (x - 2)^2 + 1 = 2x - 3 \) is a:
(a) Linear Equation
(b) Quadratic Equation
(c) Cubic Equation
(d) Bi-quadratic Equation
Answer: (b)

Question. The solution of the quadratic equation \( (x - 1)^2 - 5(x - 1) - 6 = 0 \) is:
(a) 0 and 7
(b) 0 and 1
(c) 1 and 7
(d) None of these
Answer: (a)

Question. The solution of the equation \( x^2 + 5x - (a^2 + a - 6) = 0 \) is:
(a) \( a - 2, a + 3 \)
(b) \( a + 2, a + 3 \)
(c) \( a - 2, -(a + 3) \)
(d) \( a + 2, -(a + 3) \)
Answer: (c)

Question. If the common difference of an A.P. is 5, then \( a_{18} - a_{13} \) is:
(a) 20
(b) 25
(c) 30
(d) None of these
Answer: (b)

Question. If 7 times the \( 7^{th} \) term of an A.P. is equal to 11 times its \( 11^{th} \) term, then its \( 18^{th} \) term will be :
(a) 0
(b) 1
(c) \( -1 \)
(d) None of these
Answer: (a)

Question. The fourth term from the end of the A.P., \( -11, -8, -5, \dots, 49 \) is:
(a) 30
(b) 35
(c) 37
(d) 40
Answer: (d)

Case-based MCQs

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see figure below). A competitor starts from the bucket picks up nearest potato, runs to the bucket to drop it in, and the continues in the same way until all the potatoes are in the bucket.

Question. What is the distance travelled by competitor to pick \( 1^{st} \) potato ?
(a) 10 m
(b) 16 m
(c) 22 m
(d) 48 m
Answer: (a)

Question. What is the distance travelled by competitor to pick \( 2^{nd} \) potato ?
(a) 10 m
(b) 16 m
(c) 22 m
(d) 48 m
Answer: (b)

Question. What is the distance travelled by competitor to pick \( 3^{rd} \) potato ?
(a) 10 m
(b) 16 m
(c) 22 m
(d) 48 m
Answer: (c)

Question. The common difference of A.P. involve in this problem is
(a) 2
(b) 4
(c) 6
(d) None of these
Answer: (c)

Question. The sum of 10 terms of an A.P. whose first term is 10 and common difference is 6 is :
(a) 300
(b) 370
(c) 390
(d) None of these
Answer: (b)

Very Short Answer Type Questions

Question. Find the nature of roots of the quadratic equation \( 13\sqrt{3}x^2 + 10x + \sqrt{3} = 0 \).
Answer: For the equation \( 13\sqrt{3}x^2 + 10x + \sqrt{3} = 0 \), the discriminant \( D = b^2 - 4ac = (10)^2 - 4(13\sqrt{3})(\sqrt{3}) = 100 - 156 = -56 \). Since \( D < 0 \), the equation has no real roots.

Question. Find the roots of the quadratic equation \( 6x^2 - x - 2 = 0 \).
Answer: \( 6x^2 - x - 2 = 0 \)
\( 6x^2 - 4x + 3x - 2 = 0 \)
\( 2x(3x - 2) + 1(3x - 2) = 0 \)
\( (3x - 2)(2x + 1) = 0 \)
\( x = \frac{2}{3} \) or \( x = -\frac{1}{2} \).
The roots are \( \frac{2}{3} \) and \( -\frac{1}{2} \).

Question. If the \( n^{th} \) term of an A.P., \( -1, 4, 9, 14, \dots \) is 129, then find the value of \( n \).
Answer: In the A.P. \( -1, 4, 9, 14, \dots \), \( a = -1 \) and \( d = 5 \).
Given \( a_n = a + (n - 1)d = 129 \)
\( -1 + (n - 1)5 = 129 \)
\( 5(n - 1) = 130 \)
\( n - 1 = 26 \)
\( n = 27 \).

Short Answer Type Questions-I

Question. If \( x = \frac{2}{3} \) and \( x = -3 \) are roots of the quadratic equation \( ax^2 + 7x + b = 0 \), then
Answer: Sum of roots: \( \frac{2}{3} + (-3) = -\frac{7}{3} \).
From equation, Sum \( = -\frac{7}{a} \). So, \( -\frac{7}{a} = -\frac{7}{3} \Rightarrow a = 3 \).
Product of roots: \( \left(\frac{2}{3}\right)(-3) = -2 \).
From equation, Product \( = \frac{b}{a} \). So, \( \frac{b}{3} = -2 \Rightarrow b = -6 \).

Question. For what value of \( k \), the equation \( 2x^2 + kx + 3 = 0 \) has two equal roots.
Answer: For equal roots, \( D = b^2 - 4ac = 0 \).
\( k^2 - 4(2)(3) = 0 \)
\( k^2 = 24 \)
\( k = \pm\sqrt{24} = \pm 2\sqrt{6} \).

Question. If the second term of an A.P., is 13 and the fifth term is 25, then find its \( 7^{th} \) term.
Answer: \( a + d = 13 \) and \( a + 4d = 25 \).
Subtracting equations: \( 3d = 12 \Rightarrow d = 4 \).
Substituting \( d = 4 \) in first equation: \( a + 4 = 13 \Rightarrow a = 9 \).
\( 7^{th} \) term \( a_7 = a + 6d = 9 + 6(4) = 9 + 24 = 33 \).

Short Answer Type Questions-II

Question. If the square of smaller number is 4 times the larger number and the difference between the squares of two numbers is 45, then find the numbers.
Answer: Let the larger number be \( x \) and smaller number be \( y \).
Given \( y^2 = 4x \) and \( x^2 - y^2 = 45 \).
\( x^2 - 4x = 45 \Rightarrow x^2 - 4x - 45 = 0 \)
\( (x - 9)(x + 5) = 0 \). Since \( x \) must be positive for \( y^2 = 4x \), \( x = 9 \).
\( y^2 = 4(9) = 36 \Rightarrow y = \pm 6 \).
The numbers are 9 and 6 (or 9 and -6).

Question. If the sum of the first \( n \) terms of an AP is \( 4n - n^2 \), what is the first term (that is \( S_1 \))? What is the sum of first two terms ? What is the second term ? Similarly, find the \( 3^{rd} \), the \( 10^{th} \) and the \( n^{th} \) terms.
Answer: \( S_n = 4n - n^2 \).
First term \( a_1 = S_1 = 4(1) - 1^2 = 3 \).
Sum of first two terms \( S_2 = 4(2) - 2^2 = 4 \).
Second term \( a_2 = S_2 - S_1 = 4 - 3 = 1 \).
Common difference \( d = a_2 - a_1 = 1 - 3 = -2 \).
\( 3^{rd} \) term \( a_3 = a + 2d = 3 + 2(-2) = -1 \).
\( 10^{th} \) term \( a_{10} = a + 9d = 3 + 9(-2) = -15 \).
\( n^{th} \) term \( a_n = a + (n - 1)d = 3 + (n - 1)(-2) = 3 - 2n + 2 = 5 - 2n \).

Long Answer Type Questions

Question. John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find out how many marbles they had to start with.
Answer: Let John have \( x \) marbles. Then Jivanti has \( 45 - x \) marbles.
After losing 5 marbles each, John has \( x - 5 \) and Jivanti has \( 40 - x \) marbles.
Product: \( (x - 5)(40 - x) = 124 \)
\( 40x - x^2 - 200 + 5x = 124 \)
\( -x^2 + 45x - 324 = 0 \Rightarrow x^2 - 45x + 324 = 0 \)
Factorizing: \( (x - 36)(x - 9) = 0 \).
So, they had 36 and 9 marbles respectively.