Problems on Arithmetic Geometric Progressions
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
tn = 109
n = ?
We have tn = a + (n-1) d
⇒ 109 = 4 + (n-1) 5
⇒ 109 = 4 + 5n - 5
⇒ 110 = 5n
⇒ n = 22
Q2. The 9th 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
t9 = ?
We have t9 = a + (9-1) d
= 2 + 8 X 2
= 2 + 16 = 18
Q3. If the 4th 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 t4 = 12
t12 = 60
a = ?
Now t4 = 12
⇒ a + (4-1) d = 12
⇒ a + 3d = 12 -------> (1)
t12 = 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, t5 = ?
Now t5 = 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
tn = 1089, n = ?
Now tn = 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
tn = 105, n = ?
Now tn = 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
t11 = ?
We have t11 = 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
t10 = 20
t20 = 10
tx = ?
Now, t10 = a + (10-1) d
⇒ 20 = a + 9 X (-b)
⇒ 20 = a - 9b -------------> (1)
t20 = 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
∴ tx = a + (x - 1) d
= 29 + (x - 1) (-1)
= 29 - (x - 1)
= 29 -x + 1 = 30 - x
Practice Test Exam