Commerce Exam  >  Commerce Tests  >  Mathematics (Maths) Class 11  >  Test: Special Sequences And Series - Commerce MCQ

Test: Special Sequences And Series - Commerce MCQ


Test Description

10 Questions MCQ Test Mathematics (Maths) Class 11 - Test: Special Sequences And Series

Test: Special Sequences And Series for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation. The Test: Special Sequences And Series questions and answers have been prepared according to the Commerce exam syllabus.The Test: Special Sequences And Series MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Special Sequences And Series below.
Solutions of Test: Special Sequences And Series questions in English are available as part of our Mathematics (Maths) Class 11 for Commerce & Test: Special Sequences And Series solutions in Hindi for Mathematics (Maths) Class 11 course. Download more important topics, notes, lectures and mock test series for Commerce Exam by signing up for free. Attempt Test: Special Sequences And Series | 10 questions in 10 minutes | Mock test for Commerce preparation | Free important questions MCQ to study Mathematics (Maths) Class 11 for Commerce Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Special Sequences And Series - Question 1

Let a, b, c, be in A.P. ,with a common difference d, then e1/e ,  eb/ae​, e1/a are in

Test: Special Sequences And Series - Question 2

Sum to 20 terms of the series (1.3)2 + (2.5)2 + (3.7)2 +… is:

Detailed Solution for Test: Special Sequences And Series - Question 2

Given that (1.3)2, (2.5)2 , (3.7)2 .....................20 terms
Here 1,2,3,.................are in A.P.
a = 1, d = 1
tn = a + (n-1) d = 1 + (n -1)1 = n
3,5,7 ............. are in A.P
a = 3, d = 2
tn = a + (n-1) d = 3 + (n -1)2 = 2n + 1
∴ nth term = n(2n + 1)2 = 4n3 + 4n2 + n
Sum of n terms Sn = ∑ ( 4n3 + 4n2 + n ) = 4 { n2(n+1)2}/4 + 4 {n (n+1)(2n+1)} / 6 + n(n+1)/2
= { n2(n+1)2} + 2 {n (n+1)(2n+1)} / 3 + n(n+1)/2
Sum of 20 terms S20 = 400 × 441 + 40 × 41 × 7 + 10 ×21 = 176400 + 11480 + 210
 = 188090.

Test: Special Sequences And Series - Question 3

0.2 + 0.22 + 0.222 + ……. to n terms =

Detailed Solution for Test: Special Sequences And Series - Question 3

0.2 + 0.22 + 0.222 + 0.2222 + ….
2(0.1 + 0.11 + 0.111 + 0.1111 + ….)
2/9 (0.9 + 0.99 + 0.999 + 0.9999 + ….)
2/9 (( 1 - 0.1 ) + ( 1 - 0.01 ) + ( 1 - 0.001 ) + ( 1 - 0.0001 ) + …. )
2/9 (1 - 0.1 + 1 - 0.01 + 1 - 0.001 + 1 - 0.0001 + ….)
2/9 ((1 + 1 + 1 + 1 + ….) - ( 0.1 + 0.01 + 0.001 + 0.0001 + ….))
2/9 (n - ( 0.1 + 0.01 + 0.001 + 0.0001 + ….))
2/9 (n - (( 0.1 ( 1 - 0.1n)/( 1 - 0.1))))
2/9 (n - (( 0.1 / 0.9)( 1 - 0.1n)))
2/9 (n - (1/9)(1 - 0.1n))

Test: Special Sequences And Series - Question 4

The sum of the series 32 + 62 + 92 + … + (3n)2 is:

Detailed Solution for Test: Special Sequences And Series - Question 4

32 + 62 +....
∑(3n)2
9∑n2
9n(n+1)(2n+1)/6
= 3n(n+1)(2n+1)/2

Test: Special Sequences And Series - Question 5

In how many different ways the letters of the word HEXAGON can be permuted?

Detailed Solution for Test: Special Sequences And Series - Question 5

The given word consists of 7 different letters out of which 3 are vowels and 4 are consonants.
7 different letters can be arranged in 
7! ways = 5040 ways.

Test: Special Sequences And Series - Question 6

The sum of the squares of the first 15 natural numbers is

Detailed Solution for Test: Special Sequences And Series - Question 6

Sn = 1/6n(n+1)(2n+1)
Here n = 15 so,
S15 = 1/6 × 15(15+1)(2×15+1)
S15 = 1/6 × 15 × 16(30 + 1)
S15 = 1/6 × 15 × 16 × 31
S15 = 1240

Test: Special Sequences And Series - Question 7

Sum of squares of the first n natural numbers is given by:

Detailed Solution for Test: Special Sequences And Series - Question 7

The formula to find sum of first n terms of a sequence of squares of natural numbers = n(n+1)(2n+1)/6

Test: Special Sequences And Series - Question 8

The sum of the series 13 – 23 + 33 – 43 + ……. + 93 is:

Detailed Solution for Test: Special Sequences And Series - Question 8

13 + 33 + 53 + 73 + 93 − (23 + 43 + 63 + 83)
= Sum of cubes of first 5 odd number -Sum of cubes of first 4 even number
= n2(2n2–1) − 2n2(n+1)2
= 1225 – 800
= 425

Test: Special Sequences And Series - Question 9

Sum to n terms of the series 2 + 5 + 14 + 41 +… is:

Detailed Solution for Test: Special Sequences And Series - Question 9

Sn = 2 + [2+31] + [2+31+32] + ⋯ + [2+31+32+⋯+3n-1
 = 2n+[31]+[31+32]+⋯+[31+32+⋯+3n-1]
 = 2n + t3,1 + t3,2 +⋯+t3,n-1
 = 2n+T3,n-1
 = 2n+3/4(3n−2n−1)
 =  n/2 + 3/4(3n − 1)

Test: Special Sequences And Series - Question 10

If 1, a and P are in A. P. and 1, g and P are in G. P., then

Detailed Solution for Test: Special Sequences And Series - Question 10

 2a = 1 + p and g2 = P
⇒ g2 = 2a − 1
⇒ 1 − 2a + g2 = 0

83 videos|237 docs|99 tests
Information about Test: Special Sequences And Series Page
In this test you can find the Exam questions for Test: Special Sequences And Series solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Special Sequences And Series, EduRev gives you an ample number of Online tests for practice
83 videos|237 docs|99 tests
Download as PDF
Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!