Question

Identify the 17th term of a geometric sequence where a1 = 16 and a5 = 150.06. Round the common ratio and 17th term to the nearest hundredth

Answer 1

`a_5=ra_4=r^2a_3=r^3a_2=r^4a_1\iff150.06=16r^4\implies r\approx\pm1.75`

Regardless of the sign of the common ratio, then, you will have


`a_(17)=r^(16)a_1\approx16(\pm1.75)^(16)\approx123802.31`

The actual term would be closer to
`123794` if you didn't have to round.