Insert 6 geometric means between 27 and 1 81 .

Let the 6 geometric means between 27 and 1 81 be G 1 , G 2 , G 3 , G 4 , G 5 and G 6 . Thus, 27 , G 1 , G 2 , G 3 , G 4 , G 5 , G 6 , 1 81 are in G.P. with first term a and Common ratio r ∴ a = 27 , n = 8 and a 8 = 1 81 ∵ a 8 = 1 81 ⇒ a r 7 = 1 81 ⇒ r 7 = 1 81 × a = 1 81 × 27 ⇒ r 7 = 1 3 7 ⇒ r = 1 3 ∴ G 1 = a 2 = a r = 27 1 3 = 9 G 2 = a 3 = a r 2 = 27 1 9 = 3 G 3 = a 4 = a r 3 = 27 1 27 = 1 G 4 = a 5 = a r 4 = 27 1 27 × 3 = 1 3 G 5 = a 6 = a r 5 = 27 1 27 × 9 = 1 9 G 6 = a 7 = a r 6 = 27 1 27 × 27 = 1 27 Therefore, the six GM between 27 and 1 81 are 9, 3 , 1 , 1 3 , 1 9 , 1 27 .

Insert 5 geometric means between 16 and 1 4 .

Let G 1 , G 2 , G 3 , G 4 , G 5 , be 5 geometric means between 16 and 1 4 Thus, 16 , G 1 , G 2 , G 3 , G 4 , G 5 , 1 4 are in G.P. with first term a and common ratio r . ∴ a = 16 , n = 5 and b = 1 4 ∴ Common Ratio r = b a 1 n + 1 ⇒ r = 1 16 × 4 1 6 ⇒ r = 1 2 ∴ G 1 = a 2 = a r = 16 × 1 2 = 8 G 2 = a 3 = a r 2 = 16 × 1 4 = 4 G 3 = a 4 = a r 3 = 16 × 1 8 = 2 G 4 = a 5 = a r 4 = 16 × 1 16 = 1 G 5 = a 6 = a r 5 = 16 × 1 32 = 1 2 Hence, the 5 geometric means between 16 and 1 4 are 8, 4 , 2, 1 , 1 2 .

Insert 5 geometric means between 32 9 and 81 2 .

Let the 5 geometric means between 32 9 and 81 2 be G 1 , G 2 , G 3 , G 4 and G 5 Thus, 32 9 , G 1 , G 2 , G 3 , G 4 , G 5 , 81 2 are in G.P. ∴ a = 32 9 , b = 81 2 and n = 5 ∴ Common Ratio r = b a 1 n + 1 ⇒ r = 81 × 9 2 × 32 1 6 ⇒ r = 3 2 ∴ G 1 = a r = 32 9 × 3 2 = 16 3 G 2 = a r 2 = 32 9 × 9 4 = 8 G 3 = a r 3 = 32 9 × 27 8 = 12 G 4 = a r 4 = 32 9 × 3 4 2 4 = 18 G 5 = a r 5 = 32 9 × 3 5 2 5 = 27 Hence, the five geometric means between 32 9 and 81 2 are 16 3 , 8 , 12 , 18 , 27 .

If a is the G.M. of 2 and 1 4 , find a .

Given that a is the G.M. between 2 and 1 4 . We know that Geometric Mean(GM) between a and b is a b ∴ G M = 2 × 1 4 ⇒ a = 1 2 ⇒ a = 1 2 ∴ value of a is 1 2 .