MCQs On Arithmetic Progression

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Jasmine Grover

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Arithmetic Progression (AP) is a numerical series in which the difference between any two subsequent integers is a fixed value. Arithmetic Sequence is another name for Arithmetic Progression. For instance, the natural number sequence 1, 2, 3, 4, 5, 6, is an Arithmetic Progression with a common difference of 1 between two subsequent terms (say 1 and 2). (2 -1). Even when dealing with odd and even numbers, the common difference between two consecutive words will be equal to 2.

Ques: Find the sum of the first 5 terms of the AP: 10, 6, 2…

  1. –320
  2. 512
  3. 10
  4. –960

The video below explains this:

Arithmetic Progression Detailed Video Explanation:

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Ans: (c) 10

Explanation: AP: 10, 6, 2, …

a = 10, d = - 4

Sum of first n terms = S(n) = (n/2) x [2a + (n – 1) x d]

S5 = (5/2) x [2 x (10) + (5 – 1) x (-4)]

= (5/2) x [20 + 4 x (-4)]

= (5/2) x (20 – 16)

= (5/2) x (4)

= 5 x 2

= 10

Ques: The d for the series of numbers -12, –6, 0, 6… is

  1. –2
  2. 6
  3. 8
  4. -1

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Ans: (b) 6

Explanation:

–12, –6, 0, 6,…

Let a(1) = -12, a(2) = -6, a(3) = 0, a(4) = 6

First relational d,

a(2) – a(1) = -6 – (-12) = 6

Second relational d,

a(3) – a(2) = 0 – (-6) = 6

Third relational d,

a(4) – a(3) = 6 – (0) = 6

all the d are equals to each other, hence

d = 6.

Ques: Find the number of multiples of 4 falling between 10 and 250 is:

  1. 30
  2. 15
  3. 60
  4. 20

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Ans: (c) 60

Explanation: 4’s multiples are:

12, 16, 20, 24, …

a = 12 and d = 4

2 is the remainder for 250 / 4.

250 – 2 = 248 which is divisible by 2.

12, 16…248

nth term, a(n) = 248

a(n) = a + (n − 1) x d

  • 248 = 12 + (n-1) × 4

236 / 4 = n - 1

59 = n - 1

n = 60

Ques. In the A.P. -3, -1/2, 2 …. The 11th term is

  1. 42
  2. -12
  3. 22
  4. 65

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Ans: (c) 22

Explanation: The A.P. is -3, -1/2, 2 …

Where a = – 3

d = a(2) – a(1) = (-1 / 2) - (-3)

⇒ (-1 / 2) + 3 = 5 / 2

a(n) = a + (n−1) x d

a(11) = 3 + (11-1) x (5 / 2)

a(11) = 3 + (10) x (5 / 2)

a(11) = -3 + 25

a(11) = 22

Ques: The name of the famous mathematician who is credited with discovering the total value of the very first 100 natural numbers is

  1. Gauss
  2. Pythagoras
  3. Euclid
  4. Newton

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Ans: (a) Gauss

Explanation: Gauss is the great mathematician who is credited with discovering the total value of the first 100 natural numbers.

Ques: For a given AP, a(n) = 4, n = 7, d = -4, find the value of a

  1. 51
  2. 37
  3. 2
  4. 28

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Ans: (d) 28

Explanation: d = -4, n = 7, a(n) = 4

a(n) = a + (n – 1) x d

a + (7 – 1) x (-4) = 4

a + 6 x (-4) = 4

a – 24 = 4

a = 4 + 24

a = 28

Ques: If a(17) exceeds it’s a(10) by 7. The d is:

  1. 2
  2. 4
  3. 1
  4. 3

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Ans: (c) 1

Explanation: for a given AP, the Nth term is:

a(n) = a + (n-1) x d

a(17) = a + (17−1) x d

a(17) = a + 16 x d

a(10) = a+9(d)

a(17) – a(10) = 7

(a + 16 x d) − (a + 9 x d) = 7

7 x d = 7

d = 1

Ques: 10, 7, 4, …, is an AP, what will be the 30th term of this series?

  1. 65
  2. 22
  3. -77
  4. 45

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Ans: (c) -77

Explanation: A.P. given in the question is 10, 7, 4, …

a = 10

d = a(2) – a(1) = 7 − 10 = −3

a(n) = a + (n−1) x d

a(30) = 10 + (30−1) x (−3)

a(30) = 10 + (29) x (−3)

a(30) = 10 − 87 = −77

Ques: What is the d of an AP in which a(18) – a(14) = 32?

  1. -6
  2. 1
  3. 5
  4. 8

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Ans: (d) 8

Explanation:

a(18) – a(14) = 32

a(n) = a + (n – 1) x d

a + (17) x d – (a + 13 x d) = 32

(17) x d – (13) x d = 32

(4) x d = 32

d = 8

Ques: 5, 8, 11, 14, … is an AP, what will be the a(10)?

  1. 23
  2. 12
  3. 32
  4. 95

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Ans: (c) 32

Explanation:

The given series : 5, 8, 11, 14, …

a = 5

d = 8 – 5 = 3

a(n) = a + (n – 1) x d

10th term = a(10) = a + (10 – 1) x d

= 5 + 9 x (3)

= 5 + 27

= 32

Ques: a and a(2) are -3 and 4, find the a(21) of the series.

  1. 26
  2. 95
  3. 137
  4. -43

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Ans: (c) 137

Explanation: a = -3 and a(2) = 4

a = -3

d = 4 - a = 4 - (-3) = 7

a(21)=a + (21-1) x d

= -3 + (20) x 7

= -3 + 140

= 137

Ques: For this given AP 3, 1, -1, -3, find the values of a and d is:

  1. 3, -2
  2. 0, 3
  3. 1, 2
  4. -2, 3

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Ans: (a) 3 and -2

Explanation: a = 3

d = a(2) – a(1)

⇒ 1 – 3 = -2

⇒ d = -2

Ques: Does 210 falls in the AP: 21, 42, 63, 84…? If yes, then on which term?

  1. 12th
  2. 10th
  3. 5th
  4. 7th

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Ans: (b) 10th

Explanation:

AP: 21, 42, 63, 84, …

a = 21

d = 42 – 21 = 21

a(n) = 210

a + (n – 1) x d = 210

21 + (n – 1) x (21) = 210

21 + 21 x n – 21 = 210

21 x n = 210

n = 10

Ques: The total value of the starting four multiples of 2 is:

  1. 20
  2. 65
  3. 45
  4. 30

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Ans: (a) 20

Explanation: The starting four multiples of 2 is 2, 4, 6, 8

a=2 and d=2

n=4

S(n) = (n / 2) x [2a + (n - 1) x d]

S(4) = (4 / 2) x [2 x (2) + (4 - 1) x 2]

= (2) x [4 + 6]

= (2) x [10]

= 20

Ques: On which number of term does 78 falls in the A.P. 3, 8, 13, 18, … is 78?

  1. 9th
  2. 20th
  3. 16th
  4. 4th

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Ans: (c) 16th

Explanation: Given, 3, 8, 13, 18, … is the AP.

a = 3

d = a(2) – a(1) = 8 − 3 = 5

Let a(n) term be 78.

a(n) = a + (n − 1) x d

78 = 3 + (n − 1) x 5

75 = (n − 1) x 5

(n − 1) = 15

n = 16

Ques: The first four terms of an AP whose a is 10 and d is 10 will be?

  1. 12, 24, 36, 48
  2. 8, 16, 24, 32
  3. 15, 30, 45, 60
  4. 10, 20, 30, 40

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Ans: (d) 10, 20, 30, 40

Explanation: a = 10, d = 10

a(1) = a = 10

a(2) = a1 + d = 10 + 10 = 20

a(3) = a2 + d = 20 + 10 = 30

a(4) = a3 + d = 30 + 10 = 40

Ques: In an AP, if d = -4, a = 28, n = 7, then an is:

  1. 2
  2. 1
  3. 4
  4. 5

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Ans: (c) 4

Explanation:

a(n) = a + (n - 1) x d

= 28 + (7 - 1) x (-4)

= 28 + 6 x (-4)

= 28 - 24

a(n) = 4

Ques: The missing terms in AP: __, 13, __, 3 are:

  1. 25, 5
  2. 15, 9
  3. 19, 7
  4. 18, 8

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Ans: (d)

Explanation: a2 = 13 and

a(4) = 3

The nth term of an AP;

a(n) = a + (n − 1) x d

a(2) = a + (2 - 1) x d

13 = a + d ………………. (i)

a(4) = a + (4 - 1) x d

3 = a + 3d ………….. (ii)

Subtracting equation (i) from (ii), we get

– 10 = 2 x d

d = – 5

put value of d in eq 1

13 = a + (-5)

a = 18

a(3) = 18 + (3 - 1) x (-5)

= 18 + 2 x (-5)

= 18 - 10 = 8

Ques: Starting from the last, the 20th term of the A.P. 3, 8, 13, …, 253 is:

  1. 131
  2. 123
  3. 137
  4. 158

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Ans: (d) 158

Explanation: A.P. is 3, 8, 13, …, 253

d= 5.

Starting from the end of the series,

253, 248, 243, …, 13, 8, 5

a = 253

d = 248 − 253 = −5

n = 20

a(20) = a + (20 − 1) x d

a(20) = 253 + (19) x (−5)

a(20) = 253 − 95

a(20) = 158

Ques: If the a(2) is 13 and a(5) is 25, then it’s a(7) is

  1. 23
  2. 65
  3. 15
  4. 33

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Ans: (d) 33

Explanation:

a(2) = 13

a + d = 13

a = 13 – d…. (i)

a(5) = 25

a + 4d = 25….(ii)

Substituting the value of (i) in (ii),

13 – d + 4 x d = 25

3 x d = 12

d = 4

So, a = 13 – 4 = 9

a(7) = a + 6d = 9 + 6(4) = 9 + 24 = 33

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