Assertion (A) : 5, 8, 11, 14, …… are in A.P., then $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,$ …. are also in A.P.

Reason (R) : If each term of a given A.P. is multiplied or divided by a given fixed number (other than 0), then 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.

4. Both A and R are false.

Sequence : 5, 8, 11, 14, ……

Common difference between consecutive terms = 11 - 8 = 8 - 5 = 3.

∴ 5, 8, 11, 14, …… is an A.P.

Sequence : $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,$ ….

Difference between first two terms = $4 - \dfrac{5}{2} = \dfrac{8 - 5}{2} = \dfrac{3}{2}$.

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

Since, difference between consecutive terms are equal.

∴ $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,$…… is an A.P.

∴ Assertion (A) is true.

Let an A.P. be 2, 4, 6, 8, ……. with common difference 2.

Multiplying each terms of the above A.P. by 3, we get :

6, 12, 18, 24, …….

Difference between consecutive terms = 12 - 6 = 18 - 12 = 6.

Since, difference between consecutive terms are equal.

∴ 6, 12, 18, 24, ……. is an A.P.

Thus, we can say that

If each term of a given A.P. is multiplied or divided by a given fixed number (other than 0), then resulting sequence is an A.P.

∴ Reason (R) is true.

Hence, Option 3 is the correct option.