Problems on Arithmetic Geometric Progressions
Q11. If x, x+2, 3x are in A.P. then x = ?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:(a) 2
x, x + 2, 3x are in A.P
∴ (x + 2) - x = 3x - (x + 2)
∴ x + 2 - x = 3x - x - 2
∴ 2 = 2x - 2
∴ 2x = 4
∴ x = 2
Q12. The third term of a geometrical progression is 5. The product of the first 5 terms is
(a) 22
(b) 33
(c) 44
(d) 55
Answer:(d) 55
t3 = 5
ar3-1 = 5
ar2 = 5
Product of 1st 5 terms is, a X ar X ar2 X ar3 X ar4
= a5r10
= (ar2)5
= 55
Q13. In a G.P, the first term is 4 and the common ratio is 2. The 7th term is
(a) 198
(b) 206
(c) 256
(d) 312
Answer:(c) 256
Here a = 4
r = 2
t7 = ?
tn = ar7-1
=ar6
=4 X 26
=4 X 64
=256
Q14. The sum of first 15 natural odd number is
(a) 175
(b) 225
(c) 275
(d) 315
Answer:(b) 225
The first 15 odd number are
1, 3, 5, 7, 9, 11, 13, ..... 29
Now 1 + 3 + 5 + ...... + 29
Here a = 1, n = 15, d = 2
∴ Sn = n/2 { 2a + (n-1) d}
∴ = 15/2 { 2 X 1 + (15-1) X 2 }
∴ = 15/2 { 2 + 28 }
∴ = 15 X 15
∴ = 225
Q15. Three numbers are in G.P. Their sum is 28 and product is 512. The numbers are
(a) 4, 8, 16
(b) 6, 12, 24
(c) 3, 9, 27
(d) 2, 4, 16
Answer:(a) 4, 8, 16
Let the numbers be a/r, a, ar
Now | a | + a + ar = 28 ------> (1) |
r |
And | a | . a . ar = 512 |
r |
⇒ a3 = 512
⇒ a3 = 83
⇒ a = 8
(1) ⇒ a ( | 1 | + 1 + r ) = 28 |
r |
⇒ 8 ( | 1 | + 1 + r ) = 28 |
r |
⇒ | 1 + r + r2 | = | 28 |
r | 8 |
⇒ 2r2 + 2r + 2 = 7r
⇒ 2r2 - 5r + 2 = 0
⇒ 2r2 - 4r - r + 2 = 0
⇒ 2r(r-2) - (r-2) = 0
⇒ (r-2)(2r-1) = 0
∴ r - 2 = 0 or 2r - 1 = 0
⇒ r = 2 ⇒ r = 1/2
∴ numbers are 4, 8, 16
Q16. Which term of the series 64 + 56 + 48 + ------ is zero
(a) 7
(b) 9
(c) 11
(d) 12
Answer:(b) 9
Let the nth term is zero
Here a = 64, d = 56 - 64 = -8
Now tn = a + (n-1) d
⇒ 0 = 64 + (n-1) X (-8)
⇒ 0 = 64 - 8n + 8
⇒ 8n = 72
⇒ n = 9
∴ 9th term is zero
Q17. How many 3-digit numbers are divisible by 5 ?
(a) 160
(b) 165
(c) 175
(d) 180
Answer:(d) 180
The 3-digit numbers which are divisible by 5 are
100, 105, 110, 115, ...... 995
Here a = 100
d = 105 - 100 = 5
tn = 995
n = ?
We have tn = a + (n-1) d
⇒ 995 = 100 + (n-1) 5
⇒ n-1 = (995-100)/5
⇒ n-1 = 179
⇒ n = 180
Q18. How many numbers are there between 20 and 200 which are exactly divisible by 6 ?
(a) 20
(b) 25
(c) 30
(d) 35
Answer:(c) 30
The numbers are .....
24, 30, 36, 42, .........., 198
Here a = 24
d = 30 - 24 = 6
tn = 198
n = ?
We have, tn = a + (n-1) d
⇒ 198 = 24 + (n-1) 6
⇒ n - 1 = (198 - 24) / 6
⇒ n - 1 = 29
⇒ n = 30
Q19. The first term of an A.P. is 6 and its common difference is 5. The 10th term is
(a) 39
(b) 43
(c) 51
(d) 57
Answer:(c) 51
a = 6, d = 5; t10 = ?
We have,
t10 = a + (10-1) d
= 6 + 9 X 5
= 6 + 45
= 51
Q20. What is the next number in A.P. 7, 9, 11, 13, ?
(a) 15
(b) 17
(c) 19
(d) 21
Answer:(a) 15
Here a = 7, d = 3; t5 = ?
Now,
t5 = a + (5-1) d
= 7 + 4 X 2
= 7 + 8
= 15
Practice Test Exam