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QUESTION: 1

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

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

Solution:

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.

QUESTION: 2

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

Solution:

QUESTION: 3

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

Solution:

QUESTION: 4

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

Solution:

QUESTION: 5

The first three terms of sequence when nth term tn is n^{2} – 2n are

Solution:

QUESTION: 6

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

Solution:

QUESTION: 7

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

Solution:

QUESTION: 8

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

Solution:

QUESTION: 9

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

Solution:

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

QUESTION: 10

The nth term of the series whose sum to n terms is 5n^{2} + 2n is

Solution:

It is given that the sum of n terms is

QUESTION: 11

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

Solution:

QUESTION: 12

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

Solution:

QUESTION: 13

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

Solution:

QUESTION: 14

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

Solution:

QUESTION: 15

The two arithmetic means between –6 and 14 is

Solution:

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

QUESTION: 16

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

Solution:

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 )

QUESTION: 17

The sum of n terms of an AP is 3n^{2} + 5n. A.P. is

Solution:

The sum of n terms of an A.P. = 3n^{2}+5n

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

So nth term of A.P will be Tn=3n^{2}+5n- 3(n-1)^{2}-5(n-1)= 3n^{2}+5n-3n^{2}-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,____

QUESTION: 18

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

Solution:

QUESTION: 19

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

Solution:

QUESTION: 20

The arithmetic mean between 33 and 77 is

Solution:

QUESTION: 21

The 4 arithmetic means between –2 and 23 are

Solution:

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 3^{rd} term of the AP is

Solution:

QUESTION: 23

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

Solution:

QUESTION: 24

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

Solution:

QUESTION: 25

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

Solution:

QUESTION: 26

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

Solution:

QUESTION: 27

t_{8} of the series 6, 12, 24,…is

Solution:

QUESTION: 28

t_{12} of the series –128, 64, –32, ….is

Solution:

QUESTION: 29

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

Solution:

QUESTION: 30

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

Solution:

QUESTION: 31

The last term of the series 1, –3, 9, –27 up to 7 terms is

Solution:

QUESTION: 32

The last term of the series x^{2}, x, 1, …. to 31 terms is

Solution:

QUESTION: 33

The sum of the series –2, 6, –18, …. To 7 terms is

Solution:

QUESTION: 34

The sum of the series 24, 3, 8, 1, 2, 7,… to 8 terms is

Solution:

Sum of series 24,3,8,1,2,7..to 8th term is:

Take the 1st and 2nd terms, 24,3, put them together, 243 = 3^5.

Take the 3rd and 4th terms, 8,1, put them together, 81 = 3^4.

Take the 5th and 6th terms, 2,7, put them together, 27 = 3^3.

Take the 7th and 8th terms, 0,9, put them together, 09 = 3^2.

Sum of 8 terms = 24+3+8+1+2+7+0+9 = 54

QUESTION: 35

The sum of the series to 18 terms is

Solution:

QUESTION: 36

The second term of a G P is 24 and the fifth term is 81. The series is

Solution:

QUESTION: 37

The sum of 3 numbers of a G P is 39 and their product is 729. The numbers are

Solution:

QUESTION: 38

In a G. P, the product of the first three terms 27/8. The middle term is

Solution:

QUESTION: 39

If you save 1 paise today, 2 paise the next day 4 paise the succeeding day and so on, then your total savings in two weeks will be

Solution:

If we save 1 paise today,2 paise the next day,4 pause the succeeding day and so on then you total savings in two weeks the sequence should be Geometric progression .

**1****,**** ****,****2****,****4****,**** ****8****,**** ****1****6****,**** ****.****.****.****.****.****,****1****4**** ****terms**

**First**** ****term**** ****(****a****)**** ****=**** ****1****,**

QUESTION: 40

Sum of n terms of the series 4 + 44 + 444 + … is

Solution:

S=4+44+444+-------n terms

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