If p, 1, q are in AP and p, 2, q are in GP, then which of the following statements is/are correct?

I. p, 4, q are in HP.

II. (1/p), (1/4), (1/q) are in AP.

Select the answer using the code given below.

Concept:

1. Arithmetic Progression (AP): The difference between consecutive terms is constant.

⇒ If p, a, q are in AP, then 2a = p + q.

2. Geometric Progression (GP): The ratio of consecutive terms is constant.

⇒ If p, b, q are in GP, then b² = pq.

3. Harmonic Progression (HP): Reciprocal of terms are in AP.

⇒ If p, c, q are in HP, then (1/p), (1/c), (1/q) are in AP.

Calculation:

Using AP condition: p, 1, q are in AP.

⇒ 2(1) = p + q

⇒ p + q = 2 ............(1)

Using GP condition: p, 2, q are in GP.

⇒ 2² = p × q

⇒ 4 = pq ............(2)

Now,

and 

∴  (1/p), (1/4), (1/q) are in AP and 

 p, 4 and q are in H.P.

Thus, both statements are true.

Hence, the correct answer is Option 3.