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.

a17 ≈ 123,802.31

a17 ≈ 30,707.05

a17 ≈ 19,684.01

a17 ≈ 216,654.05

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.

 a17 ≈ 123,802.31

 a17 ≈ 30,707.05

 a17 ≈ 19,684.01

 a17 ≈ 216,654.05

mallorysipp234 · Answer

@satellite73 can you help

johnweldon1993 · Answer

This one was weird....but it did come out to one of your choices lol

so to find the common ratio we have

150.06 x 16r^4

r^4 = 150.06
           ------
             16

$$\large r = \sqrt[4]{\frac{150.06}{16}}$$

Then round that...what do you get?

mallorysipp234 · Answer

I got 1.7499

johnweldon1993 · Answer

And rounded to the nearest hundredth...?

mallorysipp234 · Answer

im so confused

johnweldon1993 · Answer

Well you get 1.7499

The question says to round the ratio to the nearest hundredth....that would be 1.75  right?

mallorysipp234 · Answer

yes, but that's none of the answer choices....

johnweldon1993 · Answer

That's because we are not done yet...

johnweldon1993 · Answer

We need to find the 17th term of this sequence...

To do that...we take our first term...(16) and multiply it by the common ratio ...but raised to the 16th power

$$\large 16 \times 1.75^{16}$$

what do you get then?

mallorysipp234 · Answer

123802

johnweldon1993 · Answer

123,802.31...but yes..

so answer choice A

mallorysipp234 · Answer

@johnweldon1993 can you help me with this one too??

mallorysipp234 · Answer

@johnweldon1993