Question

Insert three geometric means between 2 and 81/8

Answer 1

Answer:

The three geometric means are 3, 9/2 and 27/4

Step-by-step explanation:

The nth term of a geometric sequence is expressed as Tn = $ar^{n-1}$ where;

a is the first term

r is the common ratio

n is the number of terms

Since we are to insert three geometric means between 2 and 81/8, the total number of terms in the sequence will be 5 terms as shown;

2, a, b, c, 81/8

a, b, and c are the 3 geometric mean to be inserted

T1 = $ar^{1-1}$ = 2

T1 = a = 2....(1)

T5= $ar^{5-1}$

T5 = $ar^{4}$ = 81/8... (2)

Dividing equation 1 by 2 we have;

$\frac{ar^{4} }{a}= \frac{\frac{81}{8} }{2}$

$r^{4} = \frac{81}{16}\\\\r = \sqrt[4]{\frac{81}{16} } \\r = 3/2$

Given a =2 and r = 3/2;

$T2=ar\\T2 = 2*3/2\\T2 = 3\\\\T3 = ar^{2} \\T3 = 2*\frac{3}{2} ^{2} \\T3 = 2*9/4\\T3 = 9/2\\\\T4 = ar^{3}\\T4 = 2*\frac{3}{2} ^{3} \\T4 = 2*27/8\\T4 = 27/4$

Therefore the three geometric means are 3, 9/2 and 27/4