Question

if the fifth term of GP is 1/3 and 9th term is 16/243 then its fourth term will be​

Answer 1

Step-by-step explanation:

Let a be the first term and r be the common ratio of the GP.

For a GP, the nth term is given by T

n

=ar

n−1

Given, T

5

=9

=>ar

5−1

=9

=>ar

4

=9 -- (1)

And,

T

12

=

243

1

=>ar

12−1

=

243

1

=>ar

11

=

243

1

-- (2)

Dividing eqn 2 by eqn 1, we get

=>r

7

=

243×9

1

=

3

5

×3

2

1

=

3

7

1

=>r=

3

1

Substituting in eqn 1, we get

a×(

3

1

)

4

=9

a=729

And T

9

=ar

9−1

=729×(

3

1

)

8

=

9

1

Answer 2

Let,

• a be the first term.

• r be the common ratio of GP.

Now, nth term of a GP is given by,

• $\bf\pink{T_n~=~a~r^{n~-~1}}$

Given,

➙ $\sf{T_5~=~a~r^{5~-~1}}$

➙ $\sf{a~r^{4}~=~\dfrac{1}{3}}--(1)$

And,

➙ $\sf{T_9~=~a~r^{9~-~1}}$

➙ $\sf{a~r^{8}~=~\dfrac{16}{243}}--(2)$

Dividing equ. (2) by equ. (1), we get

$\implies~\bf{r^4~=~\dfrac{16}{81}}$

$\implies~\bf\orange{r~=~\dfrac{2}{3}}$

Substitute the above value in equ (1), we get

$\implies~\bf\green{a~=~\dfrac{27}{16}}$

Now,

━≫ T₄ = $\bf{a~r^{3~-~1}}$

━≫ T₄ = $\bf{\dfrac{27}{16}~\times~\bigg(\dfrac{2}{3}\bigg)^{3}}$

━≫ T₄ = $\bf{\dfrac{27}{16}~\times~\dfrac{8}{27}}$

━≫ T₄ = $\bf\red{\dfrac{1}{2}}$