Question

how do you solve: What is the 7th term of the geometric sequence where a1 = 128 and a3 = 8?

Answer 1

\[\Large a_3=a_1(r)^{3-1}\]

Answer 2

\[\Large \therefore 8=128(r)^2\]
\[\Large \implies r^2={8 \over 128} = {1 \over 16}\]\[\Large \implies r=\pm\frac14\]

Answer 3

\[\Large a_n=a_1\times r^{(n-1)}\]\[\Large \implies a_7= 128\times \left( \pm \frac14 \right)^6={128 \over 4^6}\ \ \ \ ={2\cdot \cancel{64}\over \cancel{4^3}\cdot4^3}={1\over2\cdot4^2}\]\[\LARGE={1\over32}\]

Answer 4

note that even though \(r \) can be positive or negative .... \(a_7\) will always be positive

Answer 5

That was perfect thankyou!