Question

Find the general term of the GP with the third term 1 and seventh term 8.

Answer 1

In the GP, the 7th term is 8 times the 4th term. So

ar^6 = 8ar^3, or

r^3 = 8 or r = 2.

ar^4 = 48, or a = 48/r^4 = 48/16 = 3.

Answer 2

Answer:

The general term of the GP is $a_n=(2)^{\frac{3n-9}{4}}$

Step-by-step explanation:

Given : The GP with the third term 1 and seventh term 8.

To find : The general term of the GP ?

Solution :

The GP series is in the form $a,ar,ar^2,ar^3,....ar^n,...$

The third term of GP is 1.

So, $ar^2=1$ ....(1)

The seventh term of GP is 8.

So, $ar^6=8$ ....(2)

Divide (1) and (2),

$\frac{ar^6}{ar^2}=\frac{8}{1}$

$r^{6-2}=8$

$r^{4}=8$

$r=(8)^{\frac{1}{4}}$

$r=(2)^{\frac{3}{4}}$

Substitute in (1),

$a((2)^{\frac{3}{4}})^2=1$

$a(2)^{\frac{3}{2}}=1$

$a=\frac{1}{(2)^{\frac{3}{2}}}$

$a=(2)^{-\frac{3}{2}}$

The general term of the GP is the nth term of GP,

i.e. $a_n=ar^{n-1}$

Substitute the values,

$a_n=((2)^{-\frac{3}{2}})((2)^{\frac{3}{4}})^{n-1}$

$a_n=((2)^{-\frac{3}{2}})((2)^{\frac{3(n-1)}{4}})$

$a_n=(2)^{\frac{-6+3n-3}{4}}$

$a_n=(2)^{\frac{3n-9}{4}}$

The general term of the GP is $a_n=(2)^{\frac{3n-9}{4}}$