Let the five terms be a_{1}, a_{2}, a_{3}, a_{4}, a_{5}.

A = 27, B = \( \dfrac{1}{4}\)

Now, these 5 terms are between A and B.

So the GP is: A, a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, B.

So we now have 7 terms in GP with the first term being 16 and seventh being \(\dfrac{1}{4}\).

We know that, T_{n} = ar^{n–1}

Here, T_{n} = \( \dfrac{1}{4}\), a = 16 and

\( \dfrac{1}{4}\) = 16r^{7-1}

\( \dfrac{1}{(4×16)}\) = r^{6}

r = \( \dfrac{1}{2}\)

a_{1} = Ar == 8

a_{2} = Ar^{2} = 4

a_{3} = Ar^{3} = 2

a_{4} = Ar^{4} = 1

a_{5} = Ar^{5} = \( \dfrac{1}{2}\)

The five GM between 16 and \( \dfrac{1}{4}\) are 8, 4, 2, 1, \( \dfrac{1}{2}\)

Answered by Aaryan | 1 year agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Find the geometric means of the following pairs of numbers:

**(i) **2 and 8

**(ii) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).