In a GP if T3 = 18 and T6 = 486 Find:- T10

Answer:

The 10th term in the geometric progression is 29.

Step-by-step explanation:

Given: In a geometric series, [T3 = 18] and [T6 = 486].

To find: The term [T10]?

Solution:

A geometric sequence takes the form [a, ar, ar^2,...]

Where, a represents the first term, and r denotes the common ratio.

The nth term is expressed as [Tn = a * r^(n-1)]

From the information provided: [T3 = a * r^2 = 18]

And [T6 = a * r^5 = 486]

By dividing the second equation by the first:

[(a * r^5) / (a * r^2)] = 486 / 18

[r^3 = 27]

Taking the cube root provides: r = 3.

Inserting r into one of the equations allows us to solve for a.

Substituting r gives: [T3 = a * r^2 = 18]

Thus, the first term is a = 2, and the common ratio is r = 3.

The 10th term in the geometric progression is computed as:

[T10 = a * r^(10-1)]

[Thus, T10 = 29.]

Answer:

T10 = 39366

Step-by-step explanation:

In this geometric sequence, the third term equals 18, and the sixth term is 486.

We seek to determine the 10th term.

In a geometric progression, the nth term can be represented as:

[Tn = T1 * r^(n-1)]

where r signifies the common ratio, and T1 symbolizes the first term.

Given values: T3 = 18 and T6 = 486.

By solving these relationships:

Dividing the sixth term by the third term yields:

[T6 / T3 = r^3 = 486 / 18]

This results in: r = 3.

After establishing the ratio, we can input it into either term's formula:

To find T1, rearange: [T3 = T1 * r^2 = 18]

Thus dividing yields: T1 = 2.

Consequently, T10 = 39366.