# Problems on Arithmetic Geometric Progressions

#### Home⇒Quantitative Aptitude⇒ Arithmetic Geometric Progressions Questions

Q1. Which term of the A.P. 4, 9, 14, 19 ...... is 109 ?

(a) 12

(b) 19

(c) 22

(d) 27

**Answer:**(c) 22

Here a = 4

d = 9 - 4 = 14 - 9 = ....... = 5

t_{n} = 109

n = ?

We have t_{n} = a + (n-1) d

⇒ 109 = 4 + (n-1) 5

⇒ 109 = 4 + 5n - 5

⇒ 110 = 5n

⇒ n = 22

Q2. The 9^{th} term of the A.P. 2, 4, 6, 8, ...... is

(a) 16

(b) 18

(c) 20

(d) 22

**Answer:**(b) 18

Here a = 2

d = 4 - 2 = 6 - 4 = ....... = 2

t_{9} = ?

We have t_{9} = a + (9-1) d

= 2 + 8 X 2

= 2 + 16 = 18

Q3. If the 4^{th} term of an A.P. is 12 and 12th term is 60, then the first term is

(a) - 6

(b) - 2

(c) 3

(d) 5

**Answer:**(a) - 6

Here t_{4} = 12

t_{12} = 60

a = ?

Now t_{4} = 12

⇒ a + (4-1) d = 12

⇒ a + 3d = 12 -------> (1)

t_{12} = 60

⇒ a + (12 - 1) d = 60

⇒ a + 11d = 60 ------> (2)

(2) - (1) ⇒ 8d = 48

⇒ d = 6

(1) ⇒ a + 3d =12

⇒ a + 3 X 6 = 12

⇒ a + 18 = 12

⇒ a = 12 - 18 = -6

Q4. How many numbers between 100 and 500 are divisible by 4, 5, 6

(a) 5

(b) 7

(c) 9

(d) 11

**Answer:**(b) 7

L.C.M of 4, 5, 6 = 60

Between 100 and 500 each one must be divisible by 60

∴ Numbers are 120, 180, 240, 300, 360, 420, 480

∴ Required number is 7

Q5. Total number of integers between 100 and 200 which are divisible by both 2 and 3 is

(a) 9

(b) 13

(c) 15

(d) 17

**Answer:**(d) 17

L.C.M of 2, 3 = 2 X 3 = 6

Between 100 and 200 each number must be divisible by 6

∴ Numbers are 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192, 198

∴ Required number is 17

Q6. What is the next number in A.P. 3, 7, 11, 15

(a) 16

(b) 17

(c) 19

(d) 21

**Answer:**(c) 19

a = 3 , d = 7 - 3 = 4, t_{5} = ?

Now t_{5} = a + (5 - 1) d

= 3 + 4 X 4

= 3 + 16 = 19

Q7. How many odd numbered pages are there in a book of 1089 pages ?

(a) 485

(b) 512

(c) 545

(d) 598

**Answer:**(c) 545

The pages are

1, 3, 5, 7, ......., 1089

Here a = 1, d = 3 - 1 = 2

t_{n} = 1089, n = ?

Now t_{n} = a + (n-1) d

⇒ 1089 = 1 + (n - 1) 2

⇒ n - 1 = (1089 -1)/2

⇒ n - 1 = 544

⇒ n = 544 + 1 = 545

Q8. How many multiples of 5 are there between the integers 15 and 105, both inclusive ?

(a) 17

(b) 21

(c) 25

(d) 27

**Answer:**(b) 21

Multiples of 5's are

5, 10, 15, 20, ......, 105

Here a = 5, d = 10 - 5 = 5

t_{n} = 105, n = ?

Now t_{n} = a + (n-1) d

⇒ 105 = 5 + (n - 1) 5

⇒ n - 1 = (105 -5)/5

⇒ n - 1 = 20

⇒ n = 20 + 1 = 21

Q9. The first term of an A.P. is 3 and its common difference is 4. The 11th term is

(a) 39

(b) 41

(c) 43

(d) 45

**Answer:**(c) 43

Here a = 3, d = 4

t_{11} = ?

We have t_{11} = a + (11-1) d

= 3 + 10 X 4

= 3 + 40 = 43

Q10. If the 10th term of the sequence a, a-b, a-2b, a-3b, .... is 20 and its 20th term is 10, then its xth term is

(a) 20 - x

(b) 25 + x

(c) 25 - x

(d) 30 - x

**Answer:**(d) 30 - x

First term = a

d = a-b-a = -b

t_{10} = 20

t_{20} = 10

t_{x} = ?

Now, t_{10} = a + (10-1) d

⇒ 20 = a + 9 X (-b)

⇒ 20 = a - 9b -------------> (1)

t_{20} = a + (20 - 1) d

⇒ 10 = a + 19 (-b)

⇒ 10 = a - 19b ------> (2)

(1) - (2) ⇒ 10 = 10b

⇒ b = 1 ⇒ d = -1

(1) ⇒ 20 = a - 9 X 1

⇒ 20 = a - 9

⇒ a = 29

∴ t_{x} = a + (x - 1) d

= 29 + (x - 1) (-1)

= 29 - (x - 1)

= 29 -x + 1 = 30 - x