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

★ GEOMETRIC SEQUENCE ★

$a_{n} = a_{1} r^{n-1}$ 

We have, $a_{5} =150.06$
⇒ $a_{1} =16$
⇒ $a_{n} = a_{1} r^{n-1}$

Now, $a_{5} = a_{1} r^{5-1}$

So, $150.06=16r^{4}$
  ⇒ $r^{4} =\frac{150.06}{16}$      
              $=9.37875$

∴ $r=1.75$

Again, $a_{n} =a_{1} r^{n-1}$   
⇒ $a_{17} =16(1.75) ^{17-1}$ 
Thus, $a_{17} =16(1.75) ^{16}$

∴ $a_{17} = 123,802.31$