Question

in a geometric progression,2nd term is 6 and fifth term is 48,find the 10th terms?

Answer 1

a+d=6

a+4d=48

use elimination method

you will get

a and d

then put a and d in

a+10d

Answer 2

Answer:

1536

Step-by-step explanation:

Formula of nth term in G.P.=$T_n=ar^{n-1}$

Substitute n = 2

$T_2=ar^{2-1}$

$T_2=ar^1$

We are given that 2nd term is 6

So,$6=ar^1$   --1

Substitute n = 5 in the formula

$T_5=ar^{5-1}$

$T_5=ar^{4}$

we are given that 5th term is 48

$48=ar^{4}$    -2

substitute value of a from 1 into 2

$48=\frac{6}{r}r^{4}$  

$8=r^{3}$

$\sqrt[3]{8} =r$

$2=r$  

Substitute value of r in 1

$6=a(2)^1$

$3=a$

Now to find 10th term substitute n = 10 in formula of nth term

$T_{20}=3(2)^{10-1}$

$T_{20}=3(2)^{9}$

$T_{20}=1536$

Hence the 10th term is 1536