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Test Level 2: Progressions, Sequences & Series - 2 - CAT MCQ


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20 Questions MCQ Test - Test Level 2: Progressions, Sequences & Series - 2

Test Level 2: Progressions, Sequences & Series - 2 for CAT 2024 is part of CAT preparation. The Test Level 2: Progressions, Sequences & Series - 2 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 2: Progressions, Sequences & Series - 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 2: Progressions, Sequences & Series - 2 below.
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Test Level 2: Progressions, Sequences & Series - 2 - Question 1

Find the sum: 3x2 + 5x3 + 7x4 + 9x5 + .............

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 1

S = 3x2 + 5x3 + 7x4 + .............(1)
Sx = 3x3 + 5x4 + 7x5 + ...............(2)
Subtracting (2) from (1), we get
S(1 - x) = 3x2 + 2x3 + 2x4 + 2x5 + .........

Test Level 2: Progressions, Sequences & Series - 2 - Question 2

1.3 + 3.5 + 5.7 + ... (up to 20 terms) is equal to

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 2

Sum of 20 terms 


= 10 (2.6 + 41.8)
= 10 × 44.4 = 444

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Test Level 2: Progressions, Sequences & Series - 2 - Question 3

Find the sum of the series: 1 + 3 + 5 + 6 + 9 + 12 + 13 + .......... (10 terms)

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 3

In the given series, odd terms are in AP and even terms are in GP.
(1 + 5 + 9 + …….. 5 terms) + (3 + 6 + 12 + ………. 5 terms)
= 45 + 93 = 138.

Test Level 2: Progressions, Sequences & Series - 2 - Question 4

If a, b and c are in AP, then 1/bc, 1/ca and 1/ab are in

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 4

Given: a, b, c are in AP. Divide all the terms by abc, resulting sequence will also be in AP.

Test Level 2: Progressions, Sequences & Series - 2 - Question 5

Find the sum of the following series up to n terms:
3 + 6 + 10 + 16 + …

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 5

3 + 6 + 10 + 16 + …… n terms
= (1 + 2) + (2 + 4) + (4 + 6) + (8 + 8) + … n terms
= (1 + 2 + 4 + 8 + ... n terms) + (2 + 4 + 6 + 8 + … n terms)

Test Level 2: Progressions, Sequences & Series - 2 - Question 6

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

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 6

a2 – a1 = a3 – a2 = ........... = d ≠ 3m
a1, a1 + d, a1 + 2d, a1 + 3d, ….
d = 3λ + 1, 3λ + 2
a1, a1 + 1 + 3λ, a1 + 2 + 6λ, a1 + 3 + 9λ, ….
a1, a1 + 2 + 3λ (prime), a1 + 4 + 6λ, a1 + 6 + 9λ, …. 0
Hence, Infinite

Test Level 2: Progressions, Sequences & Series - 2 - Question 7

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.

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 7

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

Test Level 2: Progressions, Sequences & Series - 2 - Question 8

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 8

Test Level 2: Progressions, Sequences & Series - 2 - Question 9

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 _____.

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 9



2 - 2r = 1 + r

a = 2
Therefore, a3 + a3r3 + a3r6 + ....

So, option (2) is the answer.

Test Level 2: Progressions, Sequences & Series - 2 - Question 10

In an AP, the mth term is 1/n and the nth term is 1/m. The sum of first mn terms is

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 10

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

Test Level 2: Progressions, Sequences & Series - 2 - Question 11

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.  

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 11

Here, first term, a = 2 … (1) 
Fourth term = a + 3d = 6 ... (2)
∴ 2 + 3d = 6

⇒ 6800 × 2 × 3 = 8n + 4n2
⇒ 40,800 = 8n + 4n2
⇒ 4n2 + 8n - 40,800 = 0
⇒ n2 + 2n - 10,200 = 0
⇒ (n + 102)(n - 100) = 0
⇒ n = 100 or n = -102 (not possible)
∴ n = 100

Test Level 2: Progressions, Sequences & Series - 2 - Question 12

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

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 12

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

Test Level 2: Progressions, Sequences & Series - 2 - Question 13

Find the 5th term of an HP, if the sum of the reciprocals of the first nine terms of the harmonic progression is 90.

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 13

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)

Test Level 2: Progressions, Sequences & Series - 2 - Question 14

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?  

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 14

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).

Test Level 2: Progressions, Sequences & Series - 2 - Question 15

The 288th term of the series abbcccddddeeeeefffffff.... is  

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 15

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'.

Test Level 2: Progressions, Sequences & Series - 2 - Question 16

The number of terms between 30 and 530, which are divisible by 11, is  

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 16

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

Test Level 2: Progressions, Sequences & Series - 2 - Question 17

Find the sum of all natural numbers between 250 and 1000, which are exactly divisible by 3.

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 17

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

Test Level 2: Progressions, Sequences & Series - 2 - Question 18

In an AP, S4 = 28 and S8 = 48. Find S12.

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 18

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.

Test Level 2: Progressions, Sequences & Series - 2 - Question 19

The sum of five numbers in an AP is 30 and the sum of their squares is 220. Find the numbers.

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 19

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.

Test Level 2: Progressions, Sequences & Series - 2 - Question 20

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

Detailed Solution for Test Level 2: Progressions, Sequences & Series - 2 - Question 20

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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