Question

Let 9GM's are inserted between 100 & 3200. Then number of rational GM's are –

1

2

3

4

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Answer 1

Answer:

n=11

a=100

b=ar^10=3200

r=√2

2a,4a,8a,16a

4

Answer 2

Given:

9 Geometric means are inserted between the numbers 100 and 3200.

To Find:

The number of rational GMs are?

Solution:

The given problem can be solved using the concepts of Geometric progressions and Geometric mean.

1. The given numbers are 100 and 3200. A total of 9 terms are inserted between the two numbers. Let the 9 terms be denoted as a1, a2, a3, a4, a5, a6, a7, a8, a9.

=> The sequence after inserting geometric means are 100, a1, a2, a3, a4, a5, a6, a7, a8, a9, 3200. Let the common ratio of the given GP be r.

=> (a1/100) = (3200/a9),

=> (100 x r)/100 = (3200)/100 x r^9,

=> r^10 = 3200,

=> r = √2.

2. Therefore, the complete sequence is 100, 100√2,200, 200√2, 400, 400√2, 800, 800√2, 1600, 1600√2, 3200.

=> Total number of rational Geometric means = 200, 400, 800, and 1600.

=> Total number of rational geometrix mean terms = 4.

Therefore, the total number of rational GMs is 4. Option 4 is the correct answer.