We are given a geometric progression 10, 30, 90, ... We need to find: i. The 10th term of the sequence. ii. The sum of the first 6 terms of the sequence. iii. The geometric mean of 2430 and 21870.

1. Problem Description

We are given a geometric progression 10, 30, 90, ... We need to find:

i. The 10th term of the sequence.

ii. The sum of the first 6 terms of the sequence.

iii. The geometric mean of 2430 and

2

1

8

7

0.

2. Solution Steps

i. Finding the 10th term:

The first term, $a$, is

1

0. The common ratio, $r$, is $30/10 = 3$.

The $n$th term of a geometric progression is given by the formula:

$a_n = a \cdot r^{n-1}$

So, the 10th term is:

$a_{10} = 10 \cdot 3^{10-1} = 10 \cdot 3^9 = 10 \cdot 19683 = 196830$.

ii. Finding the sum of the first 6 terms:

The sum of the first $n$ terms of a geometric progression is given by the formula:

$S_n = \frac{a(r^n - 1)}{r - 1}$

So, the sum of the first 6 terms is:

$S_6 = \frac{10(3^6 - 1)}{3 - 1} = \frac{10(729 - 1)}{2} = \frac{10 \cdot 728}{2} = 5 \cdot 728 = 3640$.

iii. Finding the geometric mean of 2430 and 21870:

The geometric mean of two numbers $x$ and $y$ is given by:

$GM = \sqrt{x \cdot y}$

So, the geometric mean of 2430 and 21870 is:

$GM = \sqrt{2430 \cdot 21870} = \sqrt{53144100} = 7290$.

3. Final Answer

i. The 10th term is

1

9

6

8

3

0. ii. The sum of the first 6 terms is

3

6

4

0. iii. The geometric mean of 2430 and 21870 is 7290.