Question

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

Answer 1

T10=39366

Explanation:

In geometric progression, third term is 18 and 6th term is 486

we need to find out 10th term

In geometric progression , nth term is

`T_n= T_1(r)^(n-1)`

where r is the common ratio and T_1 is the first term

T3= 18 and T6= 486

`T_3= T_1(r)^(2)`

`18= T_1(r)^(2)`

`T_6= T_1(r)^(5)`

`486= T_1(r)^(5)`

Divide second equation by first equation

`(486= T_1(r)^(5))/(18= T_1(r)^(2))`

`27=r^3`

Take cube root on both sides

r= 3

`18= T_1(r)^(2)`, plug in 3 for 'r'

`18= T_1(3)^(2)`

Divide by 9 on both sides

t1= 2

So T1=2 and r=3

`T_n= T_1(r)^(n-1)`

`T_(10)= 2(3)^(9)`

T10=39366

Answer 2

The 10th term of the G.P is 29.

Explanation:

Given : In a GP if
`T_3=18` and
`T_6 = 486`.

To find : The term
`T_(10)` ?

Solution :

The geometric sequence is in the form,
`a,ar,ar^2,ar^3,...`

Where, a is the first term and r is the common ratio.

The nth term of G.P is
`T_n=ar^(n-1)`

We have given,
`T_3=18`

i.e.
`T_3=ar^(3-1)`

`18=ar^(2)` ....(1)

`T_6 = 486`

i.e.
`a_6=ar^(6-1)`

`486=ar^(5)` ....(2)

Solving (1) and (2) by dividing them,

`(486)/(18)=(ar^(5))/(ar^(2))`

`27=r^3`

`r=\sqrt[3]{27}`

`r=3`

Substitute in (1),

`18=a(3)^(2)`

`18=9a`

`a=2`

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

The 10th term, of GP is given by,

`T_(10)=2+(10-1)3`

`T_(10)=2+(9)3`

`T_(10)=2+27`

`T_(10)=29`

Therefore, The 10th term of the G.P is 29.