CBSE Class 10 Maths HOTs Arithmetic Progressions Set 05

Multiple Choice Questions

Question. The list of numbers \( -10, -6, -2, 2, \dots \) is
(a) an AP with \( d = -16 \)
(b) an AP with \( d = 4 \)
(c) an AP with \( d = -4 \)
(d) not an AP
Answer: (b)

Question. In an AP, if \( a = 3.5, d = 0, n = 101 \), then \( a_n \) will be
(a) 0
(b) 3.5
(c) 103.5
(d) 104.5
Answer: (b)

Question. Is a sequence defined by \( a_n = 2n^2 + 1 \) forms an AP?
(a) Yes
(b) Not
(c) Cannot be determined
(d) None of the above
Answer: (b)

Question. The sum of first 20 terms of an AP in which \( a = 1 \) and 20th term = 58 is
(a) 590
(b) 580
(c) 570
(d) 560
Answer: (a)

Question. The 10th term of an AP is 52 and 16th term is 82, then 32nd term of the AP is
(a) 152
(b) 159
(c) 162
(d) 156
Answer: (c)

Case Based MCQs

Kartik starts repaying a loan as first installment of ₹ 100. He increases the installment by ₹ 5 every month.

Question. AP formed from the given situation is
(a) 105, 110, 115, ......
(b) 100, 105, 110, ......
(c) 95, 100, 105, ......
(d) 110, 115, 120, ......
Answer: (b)

Question. The amount Kartik will pay in 30th installment is
(a) ₹ 265
(b) ₹ 235
(c) ₹ 255
(d) ₹ 245
Answer: (d)

Question. If Kartik pays ₹ 795, then it is
(a) 140th installment
(b) 150th installment
(c) 160th installment
(d) 170th installment
Answer: (a)

Question. Total amount paid in 13th and 17th installment is
(a) ₹ 380
(b) ₹ 300
(c) ₹ 360
(d) ₹ 340
Answer: (d)

Question. If he increases the installment by ₹ 6 every month, then the amount he will pay in 53th installment is
(a) ₹ 314
(b) ₹ 360
(c) ₹ 412
(d) ₹ 416
Answer: (c)

Short Answer Type Questions

Question. Two AP’s have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms? Why?
Answer: Yes, the difference between any two corresponding terms will be the same. Let the common difference be \( d \). The \( n \)-th terms of the two APs are \( a_n = 2 + (n-1)d \) and \( A_n = 7 + (n-1)d \). The difference \( A_n - a_n = [7 + (n-1)d] - [2 + (n-1)d] = 7 - 2 = 5 \). Since the difference is independent of \( n \), it remains constant for any two corresponding terms.

Question. Determine the AP whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20.
Answer: Let the first term be \( a \) and common difference be \( d \). Given \( a_5 = 19 \Rightarrow a + 4d = 19 \). Also \( a_{13} - a_8 = 20 \Rightarrow (a + 12d) - (a + 7d) = 20 \Rightarrow 5d = 20 \Rightarrow d = 4 \). Substituting \( d = 4 \) in the first equation: \( a + 4(4) = 19 \Rightarrow a + 16 = 19 \Rightarrow a = 3 \). Thus, the AP is 3, 7, 11, 15, ...

Question. Find the sum of all the 11 terms of an AP whose middle most term is 30.
Answer: In an AP with 11 terms, the middle most term is the 6th term (\( a_6 \)). Given \( a_6 = 30 \Rightarrow a + 5d = 30 \). The sum of 11 terms is \( S_{11} = \frac{11}{2}[2a + (11-1)d] = \frac{11}{2}[2a + 10d] = 11(a + 5d) \). Substituting \( a + 5d = 30 \), we get \( S_{11} = 11 \times 30 = 330 \).

Long Answer Type Questions

Question. An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP.
Answer: Total terms \( n = 37 \). Middle terms are 18th, 19th, and 20th. Sum: \( a_{18} + a_{19} + a_{20} = 225 \Rightarrow (a+17d) + (a+18d) + (a+19d) = 225 \Rightarrow 3a + 54d = 225 \Rightarrow a + 18d = 75 \). Sum of last three terms: \( a_{35} + a_{36} + a_{37} = 429 \Rightarrow (a+34d) + (a+35d) + (a+36d) = 429 \Rightarrow 3a + 105d = 429 \Rightarrow a + 35d = 143 \). Subtracting the two simplified equations: \( 17d = 68 \Rightarrow d = 4 \). Then \( a + 18(4) = 75 \Rightarrow a + 72 = 75 \Rightarrow a = 3 \). The AP is 3, 7, 11, 15, ...

Question. If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, then find the sum of first 10 terms.
Answer: Given \( S_6 = 36 \Rightarrow \frac{6}{2}[2a + 5d] = 36 \Rightarrow 2a + 5d = 12 \). Also \( S_{16} = 256 \Rightarrow \frac{16}{2}[2a + 15d] = 256 \Rightarrow 2a + 15d = 32 \). Subtracting the equations: \( 10d = 20 \Rightarrow d = 2 \). Substituting \( d = 2 \): \( 2a + 5(2) = 12 \Rightarrow 2a = 2 \Rightarrow a = 1 \). Now, \( S_{10} = \frac{10}{2}[2(1) + 9(2)] = 5[2 + 18] = 5 \times 20 = 100 \).

Question. Which term of the AP : 121, 117, 113, ... is its second negative term?
Answer: Here \( a = 121, d = -4 \). For negative terms, \( a_n < 0 \Rightarrow 121 + (n-1)(-4) < 0 \Rightarrow 121 - 4n + 4 < 0 \Rightarrow 125 < 4n \Rightarrow n > 31.25 \). The first negative term is the 32nd term (\( a_{32} = 121 + 31(-4) = -3 \)). The second negative term is the 33rd term (\( a_{33} = -3 + (-4) = -7 \)).

Question. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
Answer: Let terms be \( a_3 = A - 2d \) and \( a_7 = A + 2d \). Then \( (a+2d) + (a+6d) = 6 \Rightarrow 2a + 8d = 6 \Rightarrow a + 4d = 3 \). Also \( (a+2d)(a+6d) = 8 \). Substitute \( a = 3 - 4d \): \( (3-4d+2d)(3-4d+6d) = 8 \Rightarrow (3-2d)(3+2d) = 8 \Rightarrow 9 - 4d^2 = 8 \Rightarrow 4d^2 = 1 \Rightarrow d = \pm 1/2 \).
If \( d = 1/2, a = 3 - 4(1/2) = 1 \). Then \( S_{16} = \frac{16}{2}[2(1) + 15(1/2)] = 8[2 + 7.5] = 76 \).
If \( d = -1/2, a = 3 - 4(-1/2) = 5 \). Then \( S_{16} = \frac{16}{2}[2(5) + 15(-1/2)] = 8[10 - 7.5] = 20 \).
Answer: 20 or 76.

Question. Solve the equation \( -4 + (-1) + 2 + 5 + \dots + x = 437 \).
Answer: This is an AP with \( a = -4, d = 3 \). Let \( x \) be the \( n \)-th term. \( S_n = \frac{n}{2}[a + x] = 437 \) and \( x = -4 + (n-1)3 = 3n - 7 \).
Substituting \( x \): \( \frac{n}{2}[-4 + 3n - 7] = 437 \Rightarrow n(3n - 11) = 874 \Rightarrow 3n^2 - 11n - 874 = 0 \).
Solving for \( n \): \( n = \frac{11 \pm \sqrt{121 - 4(3)(-874)}}{6} = \frac{11 \pm \sqrt{121 + 10488}}{6} = \frac{11 \pm 103}{6} \).
Taking positive value, \( n = 114/6 = 19 \).
Then \( x = 3(19) - 7 = 57 - 7 = 50 \).