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.