This EduRev document offers 20 Multiple Choice Questions (MCQs) from the topic Progressions, Sequences & Series (Level - 2). These questions are of Level - 2 difficulty and will assist you in the preparation of CAT & other MBA exams. You can practice/attempt these CAT Multiple Choice Questions (MCQs) and check the explanations for a better understanding of the topic.
Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:Let a1, a2, a3... be in an A.P with a common difference, which is not a multiple of 3. The maximum number of consecutive terms which are in AP and are also prime numbers, is
Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:A radio set manufacturer produced 600 units in the third year and 700 units in the seventh year. Assuming that the production uniformly increases by a fixed number every year, find (i) the production in the first year (ii) the production in the 10th year and (iii) the total production in 7 years.
Explanation
Here, t3 = 600 and t7 = 700.
i.e. a + 2d = 600 ...........(1)
and a + 6d = 700 ...........(2)
Solving (1) and (2),
d = 25 and a = 550
(i) a, the production in the first year = 550
(ii) t10, the production in the 10th year = a + 9d = 550 + 9 (25) = 550 + 225 = 775
(iii) total production in 7 years
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Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:
Explanation
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Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:If the sum of an infinitely decreasing G.P. is 3 and the sum of squares of its terms is 9/2, then the sum of cubes of its terms is _____.
Explanation
2 - 2r = 1 + r
a = 2
Therefore, a3 + a3r3 + a3r6 + ....
So, option (2) is the answer.
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Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:In an AP, the mth term is 1/n and the nth term is 1/m. The sum of first mn terms is
Explanation
Given: am = 1/m
⇒ a + (m - 1)d = 1/n
⇒ an + mnd - nd = 1 ... (1)
am = 1/m
⇒ a + (n - 1)d = 1/m
⇒ am + mnd - md = 1 ... (2)
From (1) and (2), we get
an + mnd - nd = am + mnd - md
⇒ a(n - m) - (n - m)d = 0
⇒ a = d
Consider (1): an + mnd - nd = 1
dn + mnd - nd = 1
d = 1/mn
Hence, a = 1/mn
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Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:The first term of an arithmetic progression is 2 and the fourth term is 6. If the sum of first n terms of the progression is 6800, find the value of n.
Explanation
Here, first term, a = 2 … (1)
Fourth term = a + 3d = 6 ... (2)
∴ 2 + 3d = 6
Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:If the sums of first 8 and 19 terms of an AP are 64 and 361, respectively, then the sum of its first n terms will be
Explanation
Suppose that a and d be the first term and common difference of an AP.
S8 = 8/2 [2a + (8 - 1)d] = 64
⇒ 2a + 7d = 16 …… (1)
Also, S19 = 19/2 [2a + (19 - 1)d] = 361
⇒ 2a + 18d = 38 ..…. (2)
Subtracting (1) from (2), we get:
11d = 22
∴ d = 2
∴ 2a = 16 - 14 = 2
∴ a = 1
Sn = n/2 [2a + (n - 1)d] =n/2 [2 + (n - 1)2]
= 2n + n2 - 2n = n2
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Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:Find the 5th term of an HP, if the sum of the reciprocals of the first nine terms of the harmonic progression is 90.
Explanation
Let the harmonic progression be: 1/a, 1/(a + d), 1/(a + 2d), …
Therefore, we have a + (a + d) + (a + 2d) + … + (a + 8d)
= (9/2)[(a) + (a + 8d)]
= 9(a + 4d)
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Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:If the sum of the first 11 terms of an AP equals the sum of the first 19 terms, then what is the sum of the first 30 terms?
Explanation
S11 = S19
This means that the sum of terms from T12 to T19 is zero; which indicates that out of these 8 terms, half are positive and half are negative with same magnitude. Also, the next 11 terms have the same sum as the first 11 terms are having, but of opposite sign. This will result in the sum of the first 30 terms as 0 (zero).
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Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:The 288th term of the series abbcccddddeeeeefffffff.... is
Explanation
The series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, ...
1, 2, 3, 4, 5 and so on.
Sum of n integers starting from 1 is given by:
Sum of n integers starting from 1 is given by:
n1(n1 + 1) < 576
If n1 = 24, LHS < 576
Thus, for n1 = 23.
Thus, n1 = 24 will start the series from 277th term.
Also, n1 = 24 corresponds to 'x'.
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Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:The number of terms between 30 and 530, which are divisible by 11, is
Explanation
The required terms are: 33, 44, …, 528. They form an AP with a common difference of 11.
528 = 33 + (n - 1)11
⇒ n - 1 = 495/11 = 45
⇒ n = 46
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Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:Find the sum of all natural numbers between 250 and 1000, which are exactly divisible by 3.
Explanation
Clearly, the numbers between 250 and 1000 which are divisible by 3 are: 252, 255, 258, …, 999.
This is an AP with first term, a = 252, common difference = 3, and last term = 999.
Let there be n terms in this AP.
Then an = 999
⇒ a + (n - 1)d = 999
⇒ 252 + (n - 1) × 3 = 999
⇒ n = 250
∴ Required sum = Sn = n/2 [a+1]
= 250/2 [252 + 999] = 1,56,375
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Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:In an AP, S4 = 28 and S8 = 48. Find S12.
Explanation
Let the first term of the AP be a and the common difference be d.
S4 = 28 = (4/2)(2a + 3d) = 2(2a + 3d)
Or 2a + 3d = 14 ... (i)
S8 = 48 = (8/2)(2a + 7d) = 4(2a + 7d)
Or 2a + 7d = 12 ... (ii)
Now, subtracting (i) from (ii), we have
4d = -2
d = (-1/2) ...(iii)
Putting in (i), we get
a = (31/4)
S12 = (12/2)(2a + 11d)
= 6(2a + 11d)
= 12a + 66d
= 93 - 33
= 60
Hence, answer option 3 is correct.
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Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:The sum of five numbers in an AP is 30 and the sum of their squares is 220. Find the numbers.
Explanation
Let the numbers be:
a - 2d, a - d, a, a + d, a + 2d
Given that the sum = 30
a - 2d + a - d + a + a + d + a + 2d = 30
5a = 30
a = 6
(6 - 2d)2 + (6 - d)2 + (6)2 + (6 + d)2 + (6 + 2d)2 = 220
10d2 + 180 = 220
d2 = 4d = ±2
Hence, the numbers are 2, 4, 6, 8 and 10.
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Question for Practice Questions Level 2: Progressions, Sequences & Series - 2
Try yourself:There are 8436 steel balls, each of radius 1 cm, stacked in a pile with 1 ball at the top, 3 balls in the second layer, 6 in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile will be
Explanation
Numbers of balls in horizontal layers:
1, 3, 6, 10, …
in which each nth term is the sum of the first `n` natural numbers.
If there are m layers, then we should have
If we substitute the options one by one in the above equation, we see that m = 36 satisfies it.
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