Problems on Arithmetic Geometric Progressions
Q21. The sum of 6 + 7 + 8 + ...... + 18 = ?
(a) 132
(b) 156
(c) 178
(d) 210
Answer: (b) 156
a = 6 d = 1, tn = 18
n = ?, Sn = ?
We have, tn = a + (n-1) d
⇒ 18 = 6 + (n-1) 1
⇒ n - 1 = 12
⇒ n = 13
Sn = n/2 { 2a + (n-1) d }
= 13/2 {2 X 6 + (13-1) 1}
= 13/2 {12 + 12}
= 13 X 12
= 156
Q22. The sum of all even numbers between 100 and 200 is
(a) 6289
(b) 6750
(c) 7129
(d) 7561
Answer: (b) 6750
The numbers are 102, 104, 106, ....., 198
Here a = 102, d =2, tn = 198
n = ?, Sn = ?
tn = a + (n-1) d
⇒ 198 = 102 + (n-1) 2
⇒ n-1 = (198 - 102)/2 = 48
⇒ n = 49
∴ Sn = n/2 {2a + (n-1) d}
= 49/2 {2 X 102 + 48 X 2}
= 49/2 {300}
= 45 X 150 = 6750
Q23. 45 + 46 + 47 + ...... + 113 + 114 = ?
(a) 4322
(b) 4859
(c) 5147
(d) 5565
Answer: (d) 5565
a = 45, tn = 114
d = 1, n = ?
tn = a + (n-1) d
⇒ 114 = 45 + (n-1) 1
⇒ n - 1 = 69
⇒ n = 70
∴ Sn = n/2 {2a + (n-1) d}
= 70/2 {2 X 45 + 69 X 1}
= 35 {90 + 69}
= 35 X 159 = 5565
Q24. The sum of all natural integers from 35 to 70 is
(a) 1732
(b) 1756
(c) 1890
(d) 1930
Answer: (c) 1890
The A.P series is ....
35 + 36 + 37 + ..... + 70
Here a =35, d = 1
tn = 70
Sn = ?
We have, tn = a + (n-1) d
⇒ 70 = 35 + (n-1) 1
⇒ n - 1 = 35
⇒ n = 36
∴ Sn = 36/2 {2 X 35 + 35 X 1}
= 18 X 105 = 1890
Q25. The sum of all natural integers between 100 and 200 which are multiples of 3
(a) 4950
(b) 5782
(c) 7742
(d) 8964
Answer: (a) 4950
The numbers are :
102, 105, 108, ....... , 198
Here a = 102, d = 3, tn = 198
∴ tn = a + (n-1) d
⇒ 198 = 102 + (n-1) 3
⇒ n - 1 = (198 - 102)/3 = 32
⇒ n = 33
∴ Sn = 33/2 {2 X 102 + 32 X 3}
= 33/2 {204 + 96}
= 33/2 {300} = 33 X 150
= 4950
Q26. The 6th term of the G.P 3, 6, 11, 24 ....... is
(a) 54
(b) 69
(c) 81
(d) 96
Answer: (d) 96
a = 3 r = 6/3 = 12/6 = 2
Now t6 = ar6-1
= 3 X 25
= 3 X 32 = 96
Q27. In a G.P, the first term is 5 and the common ratio is 2. The 8th term is
(a) 525
(b) 580
(c) 640
(d) 720
Answer: (c) 640
a = 5 r = 2
∴ t8 = ar8-1 = ar7
= 5 X 27
= 5 X 128 = 640
Q28. Which term of the G.P 4, 8, 16, 32, ....... is 512
(a) 5
(b) 6
(c) 8
(d) 10
Answer: (c) 8
a = 4 r = 2
tn = 512 n = ?
Now, tn = arn-1
⇒ 512 = 4 X 2n-1
⇒ 2n-1 = 512/4 = 128
⇒ 2n-1 = 27
⇒ n - 1 = 7
⇒ n = 8
Q29. If the 4th and 9th terms of a G.P be 54 and 13122 respectively, then its 3rd term is
(a) 15
(b) 18
(c) 23
(d) 27
Answer: (b) 18
t4 = 54
⇒ ar4-1 = 54
⇒ ar3 = 54 -------> (1)
t9 = 13122
⇒ ar9-1 = 13122
⇒ ar8 = 13122 -------> (2)
(2) ÷ (1)
⇒ | ar8 | = | 13122 |
ar3 | 54 |
⇒ r5 = 243 = 35
⇒ r = 3
(1) ⇒ ar3 = 54
⇒ a.33 = 54
⇒ a = 54/27 = 2
∴ t3 = ar3-1
= ar2
= 2 X 32
= 2 X 9 = 18
Q30. If x = y then which of the following statements is true ?
(a) | x+y | = √ xy |
2 |
(b) | x+y | ‹ √ xy |
2 |
(c) | x+y | > √ xy |
2 |
(d) All of the above |
Answer:
(c) | x+y | > √ xy |
2 |
A.M > G.M
Answer:
∴ | x+y | > √ xy |
2 |
Practice Test Exam