5, 8, 11, 14, …………… are in AP.

Assertion (A): $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,$ …………… are also in AP.

Reason (R): If each term of a given A.P. is divided by the same non zero number, the resulting sequence 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.

Given, 5, 8, 11, 14, …………… are in AP.

Here, first term = 5, common difference = 8 - 5 = 11 - 8 = 3

Now, new sequence : $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,……………..$

The above sequence is found by dividing 5, 8, 11, 14, …………… the sequence by 2.

In new sequence,

Here,

Difference between second and first term = $4 - \dfrac{5}{2} = \dfrac{8 - 5}{2} = \dfrac{3}{2}$

Difference between third and second term = $\dfrac{11}{2} - 4 = \dfrac{11 - 8}{2} = \dfrac{3}{2}$

Difference between fourth and third term = $7 - \dfrac{11}{2} = \dfrac{14 - 11}{2} = \dfrac{3}{2}$

So, the common difference is same, means the given sequence is also in A.P..

So, Assertion is true.

The sequence $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,……………..$ this sequence is found by each term of the A.P. 5, 8, 11, 14, …………… is divided by 2.

If each term of a given A.P. is divided by the same non-zero number, the resulting sequence is an A.P.

So, Reason is true.

Hence, option 3 is the correct option.