Question

Hey can anyone help with number 27? this is my final!! it’ll mean a lotthanks in advance

Answer 1

27)

The given sequence is,

160, -80, 40, -20.

The first term of the sequence is a=160.

The common ratio of a sequence can be found by dividing a term by the previous term.

Hence, the common ratio is,

`\begin{gathered} r=(-80)/(160)=(-1)/(2) \\ r=(40)/(-80)=(-1)/(2) \\ r=(-20)/(40)=(-1)/(2) \end{gathered}`

So, the common ratio is -1/2.

The n th term of a geometric sequence can be expressed as,

`a_n=ar^(n-1)`

We have to find the 5th, 6th and 7th term of the given sequence.

`\begin{gathered} a_5=160*((-1)/(2))^(5-1)=160*((-1)/(2))^4=10 \\ a_6=160*((-1)/(2))^(6-1)=160*((-1)/(2))^5=-5 \\ a_7=160*((-1)/(2))^(7-1)=160*((-1)/(2))^6=2.5 \end{gathered}`

Therefore, the next three terms of the sequence 160, -80, 40, -20 is 10, -5, 2.5.

Hence, option a is correct.