Question

Q 3.3

Answer 1

Let G₁, G₂, G₃, G₄, G₅ be five geometric means between a = `3 5/9=32/9` and b = `40 1/2=81/2`

Then, `32/9`, G₁, G₂, G₃, G₄, G₅, `81/2` is a G.P.
Thus, we have
First term = a = `32/9`
7^th term of the G.P. = ar⁶ = `81/2`

⇒ `32/9` x r⁶ = `81/2`

⇒ r⁶ = `81/2xx9/32`

⇒ r⁶ = `729/64`

⇒ r⁶ = `(3/2)^6`

⇒ r = `3/2`

∴ G₁ = ar = `32/9xx3/2 = 16/3`

 G₂ = ar² = `32/9xx9/4 = 8`

G₃ = ar³ = `32/9xx27/8 = 12`

G₄ = ar⁴ = `32/9xx81/16 = 18`

G₅ = ar⁵ = `32/9xx243/32 = 27`