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Test Level 1: Progressions, Sequences & Series - 2 - CAT MCQ


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10 Questions MCQ Test Level-wise Tests for CAT - Test Level 1: Progressions, Sequences & Series - 2

Test Level 1: Progressions, Sequences & Series - 2 for CAT 2024 is part of Level-wise Tests for CAT preparation. The Test Level 1: Progressions, Sequences & Series - 2 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 1: 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 1: Progressions, Sequences & Series - 2 below.
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Test Level 1: Progressions, Sequences & Series - 2 - Question 1

Four positive numbers are in geometric progression. Their sum is 156. The sum of the first and the third numbers is 1/5th of the sum of the second and the fourth numbers. Find the first number.  

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

a, ar, ar2, ar3
a + ar + ar2 + ar3 = 156
a + ar2 = 1/5 (ar + ar3)
5a + 5ar2 = ar + ar3
5 + 5r2 = r + r3
r3 - 5r2 + r - 5 = 0
r2 (r - 5) + 1(r - 5) = 0
r = 5, r = ±1
a + 5a + 25a + 125a = 156
156a = 156
a = 1

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

If a, b, c, d, e and f are AMs between 2 and 12, then a + b + c + d + e + f is equal to  

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

As, 2, a, b, c, d, e, f, 12 are in AP
So, 2 + a + b + c + d + e + f + 12 = 8/2 (2 + 12)
2 + a + b + c + d + e + f + 12 = 4 × (2 + 12)
a + b + c + d + e + f = 4 × (2 + 12) - 2 - 12
a + b + c + d + e + f = 42

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Test Level 1: Progressions, Sequences & Series - 2 - Question 3

2 + 6 + 12 + 20 + 30 + … up to n terms equals

Detailed Solution for Test Level 1: Progressions, Sequences & Series - 2 - Question 3


Test Level 1: Progressions, Sequences & Series - 2 - Question 4

The sum of 3rd and 15th terms of an arithmetic progression is equal to the sum of 6th, 11th and 13th terms of the same progression. Which term of the given progression should necessarily be equal to zero?  

Detailed Solution for Test Level 1: Progressions, Sequences & Series - 2 - Question 4

Let the AP be a, a + d, a + 2d, …
∵ Tn = a + (n - 1)d
T3 + T15 = T6 + T11 + T13
2a + 2d + 14d = 3a + 5d + 10d + 12d
0 = a + 11d
0 = T12
Thus, the 12th term of the AP is 0.

Test Level 1: Progressions, Sequences & Series - 2 - Question 5

If the pth, qth, rth terms of an AP are x, y, z, respectively, then x(q - r) + y(r - p) + z(p - q) is equal to  

Detailed Solution for Test Level 1: Progressions, Sequences & Series - 2 - Question 5

a + (p - 1)d = x … (i)
a + (q - 1)d = y … (ii)
a + (r - 1)d = z … (iii)
Subtracting (ii) from (i), (ii) from (iii) & (iii) from (i), we get
⇒ (p - q)d = x - y

Given: x(q - r) + y(r - p) + z(p - r)

Test Level 1: Progressions, Sequences & Series - 2 - Question 6

What is the sum of all 3-digit numbers, which end with zero?

Detailed Solution for Test Level 1: Progressions, Sequences & Series - 2 - Question 6

100 + 110 + 120 + ……… + 990
= 10(10 + 11 + …… + 99)

= 10(4950 – 45) = 49,050

Test Level 1: Progressions, Sequences & Series - 2 - Question 7

If the fifth term of an AP is twice its third term, then the first term of the AP is

Detailed Solution for Test Level 1: Progressions, Sequences & Series - 2 - Question 7

t5 = 2t3
(a + 4d) = 2(a + 2d)
⇒ a = 0

Test Level 1: Progressions, Sequences & Series - 2 - Question 8

If a, b, c are in AP and x, y, z are in GP, then xb - c. yc - a. za - b equals

Detailed Solution for Test Level 1: Progressions, Sequences & Series - 2 - Question 8

Since a, b, c are in AP, therefore
b - a = c - b = d and c - a = 2d.
∴ xb - c × yc - a × za - b = x-d × y2d × z-d
= (xz)-d × y2d = y-2d × y2d = 1. (∵ y2 = xz)

Test Level 1: Progressions, Sequences & Series - 2 - Question 9

The sum of 3rd and 15th terms of an arithmetic progression is equal to the sum of 6th, 11th and 13th terms of the same progression. Which term of the given progression should necessarily be equal to zero?

Detailed Solution for Test Level 1: Progressions, Sequences & Series - 2 - Question 9

Let the AP be a, a + d, a + 2d, …
∵ Tn = a + (n - 1)d
T3 + T15 = T6 + T11 + T13
2a + 2d + 14d = 3a + 5d + 10d + 12d
0 = a + 11d
0 = T12
Thus, the 12th term of the AP is 0.

Test Level 1: Progressions, Sequences & Series - 2 - Question 10

Find the sum of all even numbers from 10 to 200 (inclusive), excluding those, which are multiples of 6.  

Detailed Solution for Test Level 1: Progressions, Sequences & Series - 2 - Question 10

10 + 12 + 14 + ------- + 200
a = 10, d = 2
Tn = 200 = 10 + (n - 1) × 2
⇒ n = 96


Sum of the numbers, which are multiple of 6:
12 + 18 + _ _ _ _ _ + 198
198 = 12 + (m - 1) 6
m = 32

Required Sum = 10,080 - 3,360 = 6,720

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