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Math Unit Test: Sequences & Series(June 20) - JEE MCQ


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15 Questions MCQ Test Daily Test for JEE Preparation - Math Unit Test: Sequences & Series(June 20)

Math Unit Test: Sequences & Series(June 20) for JEE 2024 is part of Daily Test for JEE Preparation preparation. The Math Unit Test: Sequences & Series(June 20) questions and answers have been prepared according to the JEE exam syllabus.The Math Unit Test: Sequences & Series(June 20) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Math Unit Test: Sequences & Series(June 20) below.
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Math Unit Test: Sequences & Series(June 20) - Question 1

Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3r2, then r2 - d is equal to:

Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 1

Math Unit Test: Sequences & Series(June 20) - Question 2

If tan, x, tanare in arithmetic progression and tan, y, tanare also in arithmetic progression, then |x - 2y| is equal to:

Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 2

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*Answer can only contain numeric values
Math Unit Test: Sequences & Series(June 20) - Question 3

The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit ''1'' and are multiples of 11, is __________. (in integer)


Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 3

Math Unit Test: Sequences & Series(June 20) - Question 4

Let Sn denote the sum of first n terms of an arithmetic progression. If S10 = 530, S5 = 140, then S20 - S6 is equal to:

Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 4

S10 = 530 ⇒ 10/2{2a + 9d} = 530
⇒ 2a + 9d = 106 ...(1)
and S5 = 140 ⇒ 5/2{2a + 4d} = 140
⇒ 2a + 4 d = 56 ...(2)
⇒ 5 d = 50 ⇒ d = 10 ⇒ a = 8
Now, S20 - S6 = 20/2{2a + 19 d} - 6/2{2a + 5d}
= 14a + 175 d
= (14 × 8) + (175 × 10)
= 1862

*Answer can only contain numeric values
Math Unit Test: Sequences & Series(June 20) - Question 5

Let 3, 6, 9, 12, ... upto 78 terms and 5, 9, 13, 17, ... upto 59 terms be two series. Therefore, the sum of the terms common to both the series is equal to ___. (in integer)


Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 5

For series of common terms:
a = 9, d = 12, n = 19
S19 = 19/2[2(9) + 18(12)] = 2223

Math Unit Test: Sequences & Series(June 20) - Question 6

The sum of 10 terms of the series  + ... is:

Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 6

Math Unit Test: Sequences & Series(June 20) - Question 7

Let an be the nth term of a G.P. of positive terms. If  = 200 and  = 100, then  is equal to:

Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 7

 ... (i) and
 ... (ii)
On dividing (i) by (ii), we get r = 2 and on adding (i) and (ii), we get:
a2 + a3 + a4 + a5 + ... + a200 + a201 = 300
 a1r + a2r + a3r + ... + a200r = 300
 r(a1 + a2 + ... + a200) = 300
 a1 + a+ ... + a200 = 150

Math Unit Test: Sequences & Series(June 20) - Question 8

Let a1, a2, ... an be a given A.P. whose common difference is an integer and Sn = a1 + a2 + ... + an. If a1 = 1, an = 300 and 15  n  50, then the ordered pair (Sn - 4, an - 4) is equal to

Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 8

*Answer can only contain numeric values
Math Unit Test: Sequences & Series(June 20) - Question 9

Let  be a sequence such that a1 = 1, a2 = 1 and an + 2 = 2an + 1 + an for all n  1. Then the value of 47  is equal to __________. (in integer)


Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 9

Math Unit Test: Sequences & Series(June 20) - Question 10

Let S1 be the sum of first 2n terms of an arithmetic progression. Let S2 be the sum of first 4n terms of the same arithmetic progression. If (S2 - S1) is 1000, then the sum of the first 6n terms of the arithmetic progression is equal to:

Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 10

Math Unit Test: Sequences & Series(June 20) - Question 11

If , then the remainder when K is divided by 6 is

Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 11

*Answer can only contain numeric values
Math Unit Test: Sequences & Series(June 20) - Question 12

If (r3 + 6r2 + 2r + 5) = (11!), then value of  is equal to ________. (in integer)


Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 12

{(r + 1)(r + 2)(r + 3) - 9(r + 1) + 8}
[{(r + 3)! - (r + 1)!) - 8{(r + 1)! - r!}]
= (13! + 12! - 2! - 3!) - 8(11! - 1)
= (12.13 + 12 - 8).11! - 8 + 8
= (160)(11)!
Hence,  = 160

Math Unit Test: Sequences & Series(June 20) - Question 13

The greatest positive integer k, for which 49k + 1 is a factor of the sum 49125 + 49124 + ... + 492 + 49 + 1, is

Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 13

1 + 49 + 492 + ........ 49125



Hence, k = 63

*Answer can only contain numeric values
Math Unit Test: Sequences & Series(June 20) - Question 14

The value of (0.16)log2.5 is equal to ________. (in integer)


Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 14

*Answer can only contain numeric values
Math Unit Test: Sequences & Series(June 20) - Question 15

If the sum of the first ten terms of the series  is , where m and n are co-prime numbers, then m + n is equal to ___________. (in integer)


Detailed Solution for Math Unit Test: Sequences & Series(June 20) - Question 15

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