If a, b, c are in AP; b, c, d are in GP; c, d, e are in HP, then which of the following is/are correct? 1. a, c and e are in GP 2. 1/a, 1/c, 1/e are in GP. Select the correct answer using the code given below:

From AP: 2b = a + c. From GP: c² = bd. From HP: d = 2ce/(c+e). Substituting, we get c²(c+e) = b·2ce, so c(c+e) = 2be. Also 2b = a+c gives b = (a+c)/2. Substituting: c(c+e) = (a+c)e, so c² + ce = ae + ce, giving c² = ae. Thus a, c, e are in GP. Consequently 1/a, 1/c, 1/e are also in GP (since (1/c)² = 1/(ae) = (1/a)(1/e)). Both statements are correct.