Test: Special Sequences And Series - Commerce MCQ

# Test: Special Sequences And Series - Commerce MCQ

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## 10 Questions MCQ Test - Test: Special Sequences And Series

Test: Special Sequences And Series for Commerce 2024 is part of Commerce 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.
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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.

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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

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