Question

find the first geometric sequence whose fourth term is 8 and whose common ratio is 1/2​

Answer 1

Answer:

1st term = 64

Step-by-step explanation:

last term = first term(common ratio^n-1)

8=a(1/2)^4-1

8=a(1/2)^3

8=a(1/8)

8=a/8

1/8

a=64

Check:

64 - 1st term

64(1/2) = 32 - 2nd term

32(1/2) = 16 - 3rd term

16(1/2) = 8 - 4th term ✔

64, 32, 16, 8 ✔

Answer 2

Given: 4th term of GP,$\[{T_4}\]=8$

Common ratio$=\frac{1}{2}$

To find: the first term of GP.

Solution:

Find the first term of GP.

$\[\begin{array}{l}{T_4} = 8\\ \Rightarrow {T_4} = a{r^3}\end{array}\]$

$\[ \Rightarrow 8 = a{\left( {\frac{1}{2}} \right)^3}\]$

$\[ \Rightarrow 8 = a \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\]$

$\[\begin{array}{l} \Rightarrow a = 8\times8\\ \Rightarrow a = 64\end{array}\]$

Hence, the first term of the GP is 64.