If A, B and C are three arithmetic progressions (APs) as given below :

A = 2, 4, 6, 8, …….. upto n terms

B = 3, 6, 9, 12, …… upto n terms

C = 0, 4, 8, 12, …… upto n terms, then

out of A + B, A - C, C - B and B - A which is/are A.P. ?

1. A + B

2. A - C

3. C - B

4. All are A.P.

In an A.P. a = -36, d = 18 and l = 36, then n is :

1. 10

2. 5

3. 15

4. 20

The sum of first 10 even natural numbers is :

1. 120

2. 110

3. 65

4. 120

For the given numbers $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}…………$

Assertion (A):

To find whether these terms form an A.P. or not. Express each term as the product of a natural number and $\sqrt{3 }$ i.e.,

$\sqrt{3} = 1 \times \sqrt{3}, \sqrt{12} = 2\sqrt{3}, \sqrt{27} = 3\sqrt{3}, \sqrt{48} = 4\sqrt{3}$, etc.

Reason (R): Since, for the given number difference between the consecutive term is same. It is an A.P.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.