Question

gp t9,if t3=18,t6=486

Answer 1

let the 1st term of gp be a and common ratio r .

3rd term = 18 ⇒ $ar^{2}$ = 18

6th term = 486 ⇒$ar^{5}$ = 486

∴ $ar^{5}$ / $ar^{2}$ = 486 / 18

⇒ $r^{3}$ = 27

⇒ $r = \sqrt[3]{27}$

        = 3  

and,

t3 = 18

a$r^{2}$ = 18

a$(3)^{2}$ =18

a(9) = 18

a = 18/9

a = 2

Now,

t9 = a$r^{8}$

t9 = 2$(3)^{8}$

t9 = 2 X 6561

t9 = 13122

thats the end

hpe it hlpd u ^-^"

Answer 2

Answer:

13122

Step-by-step explanation:

Refer answer written above