Question

What is the 7th term of the geometric sequence where a1 = 128 and a3 = 8?

A.) 0.03125

B.) 0.0625

C.) 0.125

D.) 0.15625

Answer 1

an = ar^(n-1)
a3 = 128r^(3 - 1) = 128r^2
128r^2 = 8
r^2 = 8/128 = 1/16
r = sqrt(1/16) = 1/4

 a7 = 128r^(7 - 1) = 128r^6 = 128(1/4)^6 = 128(1/4,096) = 0.03125

Answer 2

Answer: Option 'A' is correct.

Step-by-step explanation:

Since we have given that

$a_1=128\\\\a_3=8$

As we know that it is a geometric sequence :

$a_3=128r^{3-1}\\\\8=128r^2\\\\\frac{8}{128}=r^2\\\\0.0625=r^2\\\\\sqrt{0.0625}=r\\\\0.25=r$

And we need to find the 7th term of the geometric sequence.

$a_7=128r^{7-1}\\\\a_7=128(0.25)^6\\\\a_7=128\times 0.0002\\\\a_7=0.03125$

Hence, Option 'A' is correct.