PE 3 - Progressions | Algebra - Progressions
Thirty-one magazines are arranged from left to right in order of increasing prices. The price of each magazine differs by Rs. 2 from that of each adjacent magazine. For the price of the magazine at the extreme right a customer can buy the middle magazine and an adjacent one. Then:
- (a)
The adjacent magazine referred to is at the left of the middle magazine.
- (b)
The middle magazine sells for Rs. 36.
- (c)
The most expensive magazine sells for Rs. 64.
- (d)
None of these
Answer: Option A
Explanation :
Let the price of the cheapest magazine i.e. the one at the extreme left be x.
Difference, between the prices of two adjacent magazines is Rs. 2
∴ Price of extreme right or the costliest magazine will be x + 30 × 2 = x + 60
Now, the price of the magazine in middle (the 16th position) = x + 15 × 2 = x + 30
The price of the magazines, adjacent to the one in the middle is x + 28 or x + 32 depending on whether it is on the left or right of the middle magazine respectively.
Suppose, x + 60 = x + 28 + x + 30
∴ x + 60 = 2x + 58
∴ x = 2
And if x + 60 = x + 30 + x + 32
∴ 60 = x + 62
∴ x = −2 (which is not possible)
∴ The adjacent magazine is the one whose price is x + 28 i.e. one to the left of the middle magazine.
Hence, option (a).
Workspace:
Given the set of n numbers, n > 1, of which one is 1− 1/n, and all the others are 1. The arithmetic mean of the n numbers is
- (a)
1
- (b)
n - 1/n
- (c)
n - 1/n2
- (d)
1 - 1/n2
Answer: Option D
Explanation :
Sum of numbers = + 1 + 1 + 1 + ... (n - 1) times
= + n - 1
= n - 1/n
∴ Arithmetic mean of numbers =
=
Hence, option (d).
Workspace:
The angles of a pentagon are in arithmetic progression. One of the angles, in degrees, must be:
- (a)
108
- (b)
90
- (c)
72
- (d)
54
Answer: Option A
Explanation :
The sum of interior angles of a pentagon = 540°
Let the angles of the pentagon be a – 2d, a – d, a, a + d, a + 2d
∴ a – 2d + a – d + a + a + d + a + 2d = 540
∴ 5a = 540
∴ a = 108°
∴ One of the angles must be 108°.
Hence, option (a).
Workspace:
If xk+1 = xk + 1/2 for k = 1, 2, …, n-1 and x1 = 1, find x1 + x2 + ⋯ + xn.
- (a)
(n + 3)/2
- (b)
(n2 - 1)/2
- (c)
(n2 - n)/2
- (d)
(n2 + 3n)/4
Answer: Option D
Explanation :
xk+1 = xk + 1/2
∴ x1, x2, x3, ..., xn form an arithmetic progression with common difference d = 1/2
∵ x1 = 1, first term = a = 1
Sum of n terms of an arithmetic progression = n/2 × [2a + (n-1)d]
= n/2 × [2(1) + (n-1) 1/2]
= n/2 × [2 + n/2 - 1/2]
= n/2 × [(n + 3)/2]
= (n2 + 3n)/4
Hence, option (d).
Workspace:
Three numbers a, b, c, non-zero, form an arithmetic progression. Increasing a by 1 or increasing c by 2 results in a geometric progression. Then b equals:
- (a)
16
- (b)
14
- (c)
12
- (d)
10
Answer: Option C
Explanation :
a, b, c form an AP.
∴ 2b = a + c
Increasing a by 1 or c by 2 results in a GP
∴ b2 = (a + 1)c …(1)
and b2 = a(c + 2) …(2)
∴ (a + 1)c = a(c + 2)
∴ ac + c = ac + 2a
∴ c = 2a
Now, 2b = a + c
∴ 2b = a + 2a
∴ b = 3a/2
Putting this in (1), we get
(9a2)/4 = (a + 1)2a
∴ 9a/4 = 2a + 2
∴ 9a = 8a + 8
∴ a = 8
∴ b = 3a/2 = (3×8)/2 = 12
Hence, option (c).
Workspace:
The sum to infinity of + + + + ... is:
- (a)
1/24
- (b)
5/48
- (c)
1/16
- (d)
None of these
Answer: Option D
Explanation :
Let S = + + + + ...
∴ S1 = + + ... = =
S2 = + + + ... = =
Now, S = S1 + S2 = + =
Hence, option (d).
Workspace:
If the sum of the first ten terms of an arithmetic progression is four times the sum of the first five terms, then the ratio of the first term to the common difference is:
- (a)
1 : 2
- (b)
2 : 1
- (c)
1 : 4
- (d)
4 : 1
Answer: Option A
Explanation :
Let a be the first term and d be the common difference of the A.P.
∴10/2 × [2a + 9d] = 4 × 5/2 × [2a + 4d]
∴ 2a + 9d = 4a + 8d
∴ a/d = 1/2
Hence, option (a).
Workspace:
A saint has a magic pot. He puts one gold ball of radius 1 mm daily inside it for 10 days. If the weight of the first ball is 1 gm and if the radius of a ball inside the pot doubles every day, how much gold will the saint have gained at the beginning of 10th day?
- (a)
(230 – 69)/7 gm
- (b)
(230 + 69)/7 gm
- (c)
(230 – 71)/7 gm
- (d)
(230 + 71)/7 gm
Answer: Option C
Explanation :
We know, Weight is directly proportional to Volume which is directly proportional to cube of radius. Hence, if radius of a ball doubles, the volume will become 8 times, hence the weight will also become 8 times.
Ball 1: At the beginning of day 10, the weight of ball 1 would be increased 9 times.
∴ Weight of ball 1 at the beginning of day 10 = 1 × 89.
Ball 2: At the beginning of day 10, the weight of ball 2 would be increased 8 times.
∴ Weight of ball 1 at the beginning of day 10 = 1 × 88.
and so on.
Total weight of all the balls at the beginning of day 10 = 89 + 88 + 87 + ... + 80.
= (810 - 1)/(8 - 1) = (230 - 1)/7 gms.
Now, the weight gained = (230 - 1)/7 - 10 = (230 - 71)/7
Hence, option (c).
Workspace:
The 288th term of the sequence a, b, b, c, c, c, d, d, d, d… is
- (a)
u
- (b)
v
- (c)
w
- (d)
x
Answer: Option D
Explanation :
In the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e,...,
the first letter of the alphabet is written once, the second is written twice, and the nth letter is written n times.
∴ The number of letters written upto the nth letter is equal to the sum of the first n natural numbers given by, n(n + 1)/2
For n = 23, n(n + 1)/2 = 276 and for n = 24, n(n + 1)/2 = 300
This means the series contains 276 letters in all for the letter corresponding to n = 23 and 300 letters in all for the letter corresponding to n = 24.
∴ The letter corresponding to n = 24 will be the letter occupying the 277th to the 300th place in the series.
But, n = 24 corresponds to letter x.
∴ The 288th letter in the series is x.
Hence, option (d).
Workspace:
The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is
- (a)
-2
- (b)
-4
- (c)
-12
- (d)
8
Answer: Option C
Explanation :
Let the first term and the ratio of the Geometric Progression be a and r respectively.
∴ a + ar = 12
∴ a(1 + r) = 12 … (1)
Also, ar2 + ar3 = 48
∴ ar2(1 + r) = 48 …(2)
Dividing (2) by (1),
= = 4
∴ r2 = 4
∴ r = ± √4 = ± 2
Since the terms of the Geometric Progression are alternately positive and negative, r = −2
∴ From (1), a(1 − 2) = 12
∴ a = −12
Hence, option (c).
Workspace:
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