Problems on Arithmetic Geometric Progressions
Q31. 12 + 32 + 52 + ......... + 152 = ?
(a) 620
(b) 640
(c) 660
(d) 680
Answer: (d) 680
(12 + 22 + 32 + 42 + 52 + ....... + 152 ) - (22 + 42 + 62 + ....... + 142)
| = | 15 (15+1) (2X15+1) | - {1 X 22 + 22 X 22 + ........ + 22 X 72} |
| 6 |
| = | 15 X 16 X 31 | - 22{12 + 22 + 32 + .......... + 72} |
| 6 |
| = | 15 X 16 X 31 | - 4 X | 7 (7+1) (2X7 + 1) |
| 6 | 6 |
= 1240 - 560 = 680
Q32. 62 + 72 + ......... + 122 = ?
(a) 595
(b) 607
(c) 613
(d) 633
Answer: (a) 595
(12 + 22 + ......... + 62 + 72 + ....... + 122 ) - (12 + 22 + ....... + 52)
| = | 12 (12+1) (2X12+1) | - | 5 (5+1) (2X5+1) |
| 6 | 6 |
= 2 X 13 X 25 - 5 X 11
= 650 - 55
= 595
Q33. 13 + 23 + 33 +......... + 103 = ?
(a) 2137
(b) 2541
(c) 2863
(d) 3025
Answer: (d) 3025
| 13 + 23 + 33 +......... + n3 = { | n (n+1) | }2 |
| 2 |
| ∴ 13 + 23 + 33 +......... + 103 = { | 10 (10 + 1) | }2 |
| 2 |
= (5 X 11)2
= 552
= 3025
Q34. How many terms of the series 3, 9, 27, .......... will add up to 365 ?
(a) 3
(b) 5
(c) 7
(d) 8
Answer: (b) 5
a = 3, r = 3 , Sn = 363
| ∴ Sn = | a(rn - 1) | = | 3(3n - 1) |
| r - 1 | 3 - 1 |
| ⇒ 363 = | 3(3n - 1) |
| 2 |
| ⇒ 3n -1 = | 363 X 2 | = 242 |
| 3 |
⇒ 3n = 243
⇒ 3n = 35
⇒ n = 5
Q35. The second term of a G.P is 2/3 and the 5th term is 16/81 . Its 9th term is ?
(a) 1357
(b) 1413
(c) 1458
(d) 1566
Answer: (c) 1458
t2 = 2/3 t5 = 16/81
⇒ ar = 2/3 --------> (1) ⇒ ar4 = 16/81 --------> (2)
| (2) ÷ (1) ⇒ | ar4 | = | 16 | X | 3 |
| ar | 81 | 2 |
| ⇒ r3 = | 8 |
| 27 |
| ⇒ r3 = ( | 2 | ) 3 |
| 3 |
⇒ r = 3
(1) ⇒ a X 3 = 2/3
⇒ a = 2/9
∴ t9 = ar8 = 2/9 X 38
= 2/9 X 6561 = 1458
Q36. The sum (1 + 1/2 + 1/4 + 1/8 + .......... ∞) = ?
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (a) 2
a = 1 r = ½
| Sn = | a | = | 1 | = 2 |
| 1 - r | 1 - ½ |
Q37. If 3x, (x + 9) and (3x + 1) are in A.P then x = ?
(a) 14/3
(b) 15/6
(c) 15/7
(d) 17/4
Answer: (d) 17/4
A/Q x + 9 - 3x = (3x + 1) - (x + 9)
⇒ -2x + 9 = 2x - 8
⇒ -4x = -17
⇒ x = 17/4
Q38. The value of (13 + 23 + ......... + 153) - (1 + 2 + 3 + ..... + 15) = ?
(a) 13587
(b) 13856
(c) 14280
(d) 14665
Answer: (c) 14280
(13 + 23 + ......... + 153) - (1 + 2 + 3 + ..... + 15)
| = { | 15 (15 + 1) | }2 - | 15 (15 + 1) |
| 2 | 2 |
| = ( | 15 X 16 | )2 - | 15 X 16 |
| 2 | 2 |
= 1202 - 120 = 14400 - 120 = 14280
Q39. The value of 12 + 22 + ......... + 102 = 385 then the value of 22 + 42 + ...... + 202 is
(a) 1421
(b) 1435
(c) 1540
(d) 1598
Answer: (c) 1540
22 + 42 + ...... + 202
= (1 X 22 + 22 X 22 + 22 X 3 2 + ........ + 22 X 102 )
= 22 (12 + 22 + 32 + ......+ 102)
= 4 X 385 = 1540
Q40. A club consists of members whose ages are in A.P, the common difference being 3 months. If the youngest member of the club is just 7 years old and the sum of the ages of all the members is 250 years, then the number of members in the club is
(a) 21
(b) 25
(c) 26
(d) 31
Answer: (b) 25
The ages of the members are ....
7y, 7y3m, 7y6m, 7y9m ........
i.e 7y, 7(1/4)y, 7(1/2)y, .........
Here a = 7 d = 7(1/4) - 7 = 1/4
Sn = 250
Now Sn = n/2 {2a + (n-1) d}
⇒ 250 = n/2 {2 X 7 + (n-1) ¼ }
⇒ 500 = n {14 + (n/4) - ¼ }
⇒ 2000 = n X (55 + n)
⇒ 2000 = 55n + n2
⇒ n2 + 55n - 2000 = 0
⇒ n2 + 80n - 25n - 2000 = 0
⇒ n(n + 80) - 25(n + 80) = 0
⇒ (n + 80) (n - 25) = 0
∴ n + 80 = 0 or n - 25 = 0
⇒ n = - 80 ⇒ n = 25
But n ≠ -80
∴ n = 25
Practice Test Exam