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NDA 1 2025 Maths, question 6

Topic: Algebra

Q6

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

I. p,4,qp,4,q are in HP.

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

Select the answer using the code given below.

  1. (a) I only
  2. (b) II only
  3. (c) Both I and IICorrect answer
  4. (d) Neither I nor II

Solution.

Answer: (c) Both I and II

  1. AP gives p+q=2p + q = 2 and GP gives pq=4pq = 4.
  2. So 1p+1q=p+qpq=12=2×14\frac{1}{p} + \frac{1}{q} = \frac{p + q}{pq} = \frac{1}{2} = 2 \times \frac{1}{4}, which means 1/p,1/4,1/q1/p, 1/4, 1/q are in AP: statement II.
  3. Numbers whose reciprocals are in AP are in HP, so p,4,qp, 4, q are in HP: statement I.

Trap: No real p,qp, q satisfy both conditions (t2−2t+4=0t^{2} - 2t + 4 = 0 has negative discriminant); the algebra still holds for the complex roots 1±i31 \pm i\sqrt{3}.