Question

If the first term of a GP is 486 and the common ratio is (1/3), the tenth term is ?

Answer 1

Answer:

$t_{10}=\frac{2}{81}$

Step-by-step explanation:

If the first term of a GP is 486 and the common ratio is (1/3), the tenth term is ?

Formula used:

The n th term of the G.P

$a,ar,ar^2.........is\:t_n=ar^{n-1}$

Given:

First term, a=486

common ratio, $r=\frac{1}{3}$

To find :

$t_{10}$

Now,

$t_n=ar^{n-1}$

$t_{10}=ar^{10-1}$

$t_{10}=ar^{9}$

$t_{10}=(486)(\frac{1}{3})^{9}$

$t_{10}=(486)(\frac{1}{3^9})$

$t_{10}=(2*243)(\frac{1}{3^9})$

$t_{10}=(2*3^5)(\frac{1}{3^9})$

$t_{10}=2(\frac{1}{3^4})$

$t_{10}=2(\frac{1}{81})$

$t_{10}=\frac{2}{81}$

Answer 2

Geometric progression : A sequence in which the next term is found by Multiplying with some number.

General form :

If a is the first term, and r is the common ratio., GP will be a, ar, ar^2,........

n-th term :

Term n is given by ar^(n-1)

Solution :

First term of the Geometric progression (a) = 486

Common ratio (r) = (1/3)

We are required to find 10th term

Now,

Term 10

⇒a × r^(10-1)

⇒a × r^9

⇒(486) × (1/3)^9

Simplifying this ;

$486 = 2 \times {3}^{5}$

So,

$\: T_{10} = \frac{486}{3 ^{9} } = \frac{2 \times {3}^{5} }{ {3}^{9} } \\ \\ T_{10} = \: \frac{2}{ {3}^{(9 - 5)} } = \frac{2}{ {3}^{4} } \\ \\ T_{10} = \frac{2}{81}$

Therefore, The 10th term of the Geometric Progression is 2/81.