Problems on Arithmetic Geometric Progressions
Q41. A man arranges to pay off a debt of Rs. 3600 by 40 annual instalments which are in A.P. When 30 of the instalments are paid, he dies leaving one third of the debt unpaid. The value of the 10th instalment is
(a) 55
(b) 57
(c) 63
(d) 69
Answer: (d) 69
Total debt = Rs. 3600
Debt paid = Rs. (3600 X (2/3)) = Rs. 2400 paid in 30 instalments
Let the instalment be .....
a, a+d, a+2d, .......
S30 = 2400
⇒ 30/2 {2a + (30 - 1) d} = 2400
⇒ 2a + 29d = 160 --------> (1)
S40 = 3600
⇒ 40/2{2a + (40 - 1) d} = 3600
⇒ 2a + 39d = 180 --------> (2)
(2) - (1) ⇒ 10d = 20
⇒ d = 2
(1) ⇒ 2a + 29 X 2 = 160
⇒ 2a = 160 - 58
⇒ a = 51
∴ T10 = a + (10 - 1) d
= 51 + 9 X 2 = 51 + 18 = 69
Q42. Sonia purchases NSC every year whose value exceeds the previous years purchase by Rs. 400. After 8 years, she finds that she has purchased NSC whose total face value is Rs. 48000. What is the face value of the certificates purchased by her in the 1st year ?
(a) 4512
(b) 4550
(c) 4600
(d) 4657
Answer: (c) 4600
Let the required value be Rs. a
d = 400, n = 8, Sn = 48000
Sn = n/2 {2a + (n - 1) d}
⇒ 48000 = 8/2 {2a + 7 X 400 }
⇒ 48000 / 4 = 2a + 2800
⇒ 12000 - 2800 = 2a
⇒ a = 4600
Q43. A man has to pay Rs. 450 in yearly instalments, each instalment being less than the earlier one by Rs. 10. The amount of the fist instalment is Rs. 100. In what time, the entire amount will be paid ?
(a) 3 years
(b) 4 years
(c) 6 years
(d) 7 years
Answer: (c) 6 years
Let the required time be n years
The A.P is 100, 90, 80, .......
Now a = 100
d = -10
n = ?
Sn = 450
∴ sn = n/2 {2a + (n - 1) d}
⇒ 450 = n/2 {2 X 100 + (n-1) (-10)}
⇒ 900 = n {200 - 10n + 10}
⇒ 900 = n {210 - 10n}
⇒ 900 = 210n - 10n2
⇒ 10n2 - 210n + 900 = 0
⇒ n2 - 21n + 90 = 0
⇒ n2 - 15n - 6n + 90 = 0
⇒ n(n - 15) - 6(n - 15) = 0
⇒ (n - 15) (n - 6) = 0
∴ n - 15 = 0 n - 6 = 0
⇒ n = 15 n = 6
But n ≠ 15
∴ n = 6 years
Q44. How many terms there in the A.P 3, 9, 15, 21, ....... 57
(a) 7
(b) 8
(c) 9
(d) 10
Answer: (d) 10
a = 3 d = 9 - 3 = 15 - 9 = 6
tn = 57 n = ?
We have
tn = a + (n - 1) d
⇒ 57 = 3 + (n - 1) 6
⇒ n - 1 = (57 - 3)/6 = 9
⇒ n = 9 + 1 = 10
Q45. How many terms there in the G.P 3, 9, 27, ....... 729
(a) 5
(b) 6
(c) 7
(d) 8
Answer: (b) 6
a = 3 r = 9/3 = 3
tn = 729 n = ?
We have
tn = arn-1
⇒ 729 = 3. 3n-1
⇒ 729/3 = 3n-1
⇒ 243 = 3n-1
⇒ 3n-1 = 35
⇒ n - 1 = 5
⇒ n = 6
Q46. The 14th term of the A.P 14, 9, 4, -1, -6, ...... is
(a) -5
(b) -32
(c) -51
(d) -59
Answer: (c) -51
a = 14 d = 9 - 14 = -5
∴ t14 = a + (14 - 1) d
= 14 + 13 X (-5)
= 14 - 65
= -51
Q47. The 101 th term of the A.P 4, 4½, 5, 5½, 6, ....... is
(a) 41
(b) 49
(c) 54
(d) 61
Answer: (c) 54
a = 4 d = 4½ - 4 = ½
∴ t101 = a + (101 - 1) d
= 4 + 100 X ½
= 4 + 50
= 54
Q48. Which term of the A.P 2, 7, 12, 17 ........ is 87
(a) 18
(b) 20
(c) 21
(d) 24
Answer: (a) 18
a = 2 d = 5
tn = 87 n = ?
tn = a + (n - 1) d
⇒ 87 = 2 + (n - 1) 5
⇒ (n - 1) = (87 - 2)/5 = 17
⇒ n = 18
Q49. Find the sum of the series 3 + 7 + 11 + 15 + ......... + 39
(a) 160
(b) 172
(c) 184
(d) 210
Answer: (d) 210
a = 3 d = 4
tn = 39 n = ?
tn = a + (n - 1) d
⇒ 39 = 3 + (n - 1) 4
⇒ n - 1 = (39 - 3)/4 = 9
⇒ n = 10
∴ Sn = n/2 {2a + (n - 1) d}
= 10/2 {2 X 3 + (10 - 1) X 4}
= 5 {6 + 36}
= 5 X 42 = 210
Q50. The population of bacteria culture doubles every 2 minutes. How many minutes will it take for the population to grow from 100 to 51200 bacteria
(a) 14
(b) 15
(c) 20
(d) 22
Answer: (c) 20
The series is .....
100, 200, 400, 800, .....51200
Here a = 100 r = 200/100 = 2
tn = 51200 n = ?
We Have tn = arn-1
⇒ 51200 = 100 X 2n-1
⇒ 2n-1 = 51200/100 = 512
⇒ 2n-1 = 29
⇒ n-1 = 9
⇒ n = 10
∴ Time Taken = 2 X 10 min = 20 minutes
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