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Test: Arithmetic And Geometric Progressions - 1 - CA Foundation MCQ


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30 Questions MCQ Test - Test: Arithmetic And Geometric Progressions - 1

Test: Arithmetic And Geometric Progressions - 1 for CA Foundation 2024 is part of CA Foundation preparation. The Test: Arithmetic And Geometric Progressions - 1 questions and answers have been prepared according to the CA Foundation exam syllabus.The Test: Arithmetic And Geometric Progressions - 1 MCQs are made for CA Foundation 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Arithmetic And Geometric Progressions - 1 below.
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Test: Arithmetic And Geometric Progressions - 1 - Question 1

Choose the most appropriate option ( a ), ( b ) , ( c ) or (d)

The nth element of the sequence 1, 3, 5, 7,….…..Is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 1

Let a be the first term and d be the common difference.
Given series is in AP.

a = 1
d = 3 - 1 = 2

We know that nth term of an Ap an = a + (n - 1) * d

= 1 + (n - 1) * 2
= 1 + 2n - 2
= 2n - 1.

Test: Arithmetic And Geometric Progressions - 1 - Question 2

The nth element of the sequence –1, 2, –4, 8 ….. is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 2

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Test: Arithmetic And Geometric Progressions - 1 - Question 3

The arithmetic mean between a and 10 is 30, the value of ‘a’ should be

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 3

Correct answer is option C) 50.

Finding the value of 'a':
- Given: The arithmetic mean between 'a' and 10 is 30.
- Formula for arithmetic mean: (a + b) / 2
- Plug in the given values: (a + 10) / 2 = 30

Calculating 'a':
- Multiply both sides by 2: a + 10 = 60
- Subtract 10 from both sides: a = 50
Answer:
- The value of 'a' is 50 (Option C).

Test: Arithmetic And Geometric Progressions - 1 - Question 4

–5, 25, –125 , 625, ….. can be written as

Test: Arithmetic And Geometric Progressions - 1 - Question 5

The first three terms of sequence when nth term tn is n2 – 2n are

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 5

Correct answer is option A) -1, 0, 3

To find the first three terms of the sequence, we need to substitute n = 1, n = 2, and n = 3 into the nth term formula, t_n = n^2 - 2n.
First term (n = 1):
- t1 = (1)^2 - 2(1)
- t1 = 1 - 2
- t1 = -1

Second term (n = 2):
- t2 = (2)^2 - 2(2)
- t2 = 4 - 4
- t2 = 0

Third term (n = 3):
- t3 = (3)^2 - 2(3)
- t3 = 9 - 6
- t3 = 3

So, the first three terms of the sequence are:
- -1, 0, 3
Therefore, the correct answer is option a (-1, 0, 3).

Test: Arithmetic And Geometric Progressions - 1 - Question 6

Which term of the progression –1, –3, –5, …. Is –39

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 6

Correct answer is option B) 20th

Finding the term position:

  • Given an arithmetic progression: –1, –3, –5, ...
  • Common difference (d) between terms is: –3 - (-1) = -2
  • First term (a) is: –1
  • Term we are looking for is: -39
  • Formula for the nth term of an arithmetic progression: a_n = a + (n-1)d
  • Substitute the values: -39 = -1 + (n-1)(-2)
  • Solve for n: -39 + 1 = (n-1)(-2) => -38 = -2(n-1) => 19 = n-1 => n = 20

Answer: The 20th term of the progression is -39.

Test: Arithmetic And Geometric Progressions - 1 - Question 7

The value of x such that 8x + 4, 6x – 2, 2x + 7 will form an AP is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 7

Correct answer is option C) 15/2

To find the value of x such that 8x + 4, 6x - 2, and 2x + 7 form an Arithmetic Progression (AP), we can follow these steps:
Step 1: Identify the terms in the AP
- First term (a1) = 8x + 4
- Second term (a2) = 6x - 2
- Third term (a3) = 2x + 7

Step 2: Use the property of AP
In an arithmetic progression, the difference between consecutive terms is constant. Therefore, the difference between the second term and the first term should be equal to the difference between the third term and the second term.
(a2 - a1) = (a3 - a2)

Step 3: Substitute the terms and solve for x
- (6x - 2) - (8x + 4) = (2x + 7) - (6x - 2)
- (-2x - 6) = (9 - 4x)
- (-2x + 4x) = (9 + 6)
- 2x = 1
- x = 15/2
So, the value of x is 15/2, which is option (c) 15/2.

Test: Arithmetic And Geometric Progressions - 1 - Question 8

The mth term of an A. P. is n and nth term is m. The r th term of it is

Test: Arithmetic And Geometric Progressions - 1 - Question 9

An infinite GP has first term x and sum 5, then x belongs to

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 9

We know that, the sum of infinite terms of GP is

Test: Arithmetic And Geometric Progressions - 1 - Question 10

The nth term of the series whose sum to n terms is 5n2 + 2n is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 10

It is given that the sum of n terms is

Test: Arithmetic And Geometric Progressions - 1 - Question 11

The 20th term of the progression 1, 4, 7, 10.................is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 11

Correct answer is option A) 58

To find the 20th term of the given arithmetic progression, we can use the formula:

nth term (Tn) = a + (n - 1) * d
where:
- a is the first term
- n is the position of the term in the sequence
- d is the common difference between consecutive terms

For the given progression:
- a = 1 (first term)
- n = 20 (we want to find the 20th term)
- d = 4 - 1 = 3 (common difference)

Applying the formula:
T20 = 1 + (20 - 1) * 3
- T20 = 1 + 19 * 3
- T20 = 1 + 57
- T20 = 58

So, the 20th term of the progression is 58.

Test: Arithmetic And Geometric Progressions - 1 - Question 12

The last term of the series 5, 7, 9,….. to 21 terms is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 12
  • The series given is an arithmetic progression (AP) where the first term (a) is 5 and the common difference (d) is 2.
  • The formula for the nth term of an AP is: nth term = a + (n - 1) * d.
  • For 21 terms, substitute a = 5, d = 2, and n = 21: nth term = 5 + (21 - 1) * 2.
  • This simplifies to: 5 + 20 * 2 = 5 + 40 = 45.
  • Thus, the last term of the series is 45.
Test: Arithmetic And Geometric Progressions - 1 - Question 13

The last term of the A.P. 0.6, 1.2, 1.8,… to 13 terms is

Test: Arithmetic And Geometric Progressions - 1 - Question 14

The sum of the series 9, 5, 1,…. to 100 terms is

Test: Arithmetic And Geometric Progressions - 1 - Question 15

The two arithmetic means between –6 and 14 is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 15

Let the terms be – 6, a, b 14

a = – 6

T4 = a + 3d = 14

– 6 + 3d = 14

3d = 20

d = 20/3

a = – 6 + 20/3 = 2/3

b = 2/3 + 20/3 = 22/3

Test: Arithmetic And Geometric Progressions - 1 - Question 16

The sum of three integers in AP is 15 and their product is 80. The integers are

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 16

let a - d , a , a + d are three terms of an AP

according to the problem given,

sum of the terms = 15

a - d + a + a + d = 15

3a = 15

a = 15 / 3 

a = 5

product = 80

( a - d ) a ( a + d ) = 80

( a² - d² ) a = 80

( 5² - d² ) 5 = 80

25 - d² = 80 /5

25 - d² = 16

- d² = - 9

d² = 3²

d = ± 3

Therefore,

a = 5 , d = ±3

required 3 terms are

a - d = 5 - 3 = 2

a = 5

a+ d = 5 + 3 = 8

( 2 , 5 , 8 ) or ( 8 , 5 , 2 )

Test: Arithmetic And Geometric Progressions - 1 - Question 17

The sum of n terms of an AP is 3n2 + 5n. A.P. is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 17

The sum of n terms of an A.P. = 3n2+5n

then sum of n-1 terms= 3(n-1)2+5(n-1)

So nth term of A.P will be Tn=3n2+5n- 3(n-1)2-5(n-1)= 3n2+5n-3n2-3+6n-5n+5=6n+2

in this way by putinng values of n= 1,2,3,4,5,6,7,_______n-1, n

we will get the AP as 8,14,20,26,32,38,44,____

Test: Arithmetic And Geometric Progressions - 1 - Question 18

The number of numbers between 74 and 25556 divisible by 5 is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 18



Test: Arithmetic And Geometric Progressions - 1 - Question 19

The pth term of an AP is (3p – 1)/6. The sum of the first n terms of the AP is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 19

Test: Arithmetic And Geometric Progressions - 1 - Question 20

The arithmetic mean between 33 and 77 is

Test: Arithmetic And Geometric Progressions - 1 - Question 21

The 4 arithmetic means between –2 and 23 are

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 21

a=-2 n=6

given : t6= 23

23= a+5d

23= -2+5d

d= 25/5

d = 5

 

 

t2= a+d= -2+5=3

t3= 8

t4= 13

t5= 18

Test: Arithmetic And Geometric Progressions - 1 - Question 22

The first term of an A.P is 14 and the sums of the first five terms and the first ten terms are equal is magnitude but opposite in sign. The 3rd term of the AP is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 22

Test: Arithmetic And Geometric Progressions - 1 - Question 23

The sum of a certain number of terms of an AP series 8, 6, 4, …… is -52. The number of terms is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 23

Test: Arithmetic And Geometric Progressions - 1 - Question 24

The 1st and the last term of an AP are –4 and 146. The sum of the terms is 7171. The number of terms is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 24

Test: Arithmetic And Geometric Progressions - 1 - Question 25

The sum of the series 3 ½ + 7 + 10 ½ + 14 + …. To 17 terms is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 25

Test: Arithmetic And Geometric Progressions - 1 - Question 26

The 7th term of the series 6, 12, 24,……is

Detailed Solution for Test: Arithmetic And Geometric Progressions - 1 - Question 26

 

Test: Arithmetic And Geometric Progressions - 1 - Question 27

t8 of the series 6, 12, 24,…is

Test: Arithmetic And Geometric Progressions - 1 - Question 28

t12 of the series –128, 64, –32, ….is

Test: Arithmetic And Geometric Progressions - 1 - Question 29

The 4th term of the series 0.04, 0.2, 1, … is

Test: Arithmetic And Geometric Progressions - 1 - Question 30

The last term of the series 1, 2, 4,…. to 10 terms is

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