Question

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

Answer 1

Answer:

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

Step-by-step 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,

$\frac{486}{18}=\frac{ar^{5}}{ar^{2}}$

$27=r^3$

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

$r=3$

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.

Answer 2

Answer:

T10=39366

Step-by-step 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

$\frac{486= T_1(r)^{5}}{18= T_1(r)^{2}}$

$27=r^3$

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