Question

A sequence has a first term of 2 and a constant ratio of . What are the first five terms of the sequence?

Answer 1

The first five terms are 2,
`-(4)/(3)` ,
`(8)/(9)` ,
`-(16)/(27)` ,
`(32)/(81)`

Explanation:

In the geometric sequence, there is a constant ratio between each two consecutive numbers

Examples:

5, 10, 20, 40, 80, ………………………. (×2)

5000, 1000, 200, 40, …………………………(÷5)

General term (nth term) of a Geometric sequence is:

a1 = a, a2 = ar, a3 = ar², a4 = ar³, ..........

`an=ar^(n-1)`, where

a is the first term

r is the constant ratio between each two consecutive terms

Let us solve the question

∵ A sequence has a first term of 2 and a constant ratio of
`-(2)/(3)`

∴ This is a geometric sequence, where

• a = 2
• r =
  `-(2)/(3)`

→ We need to find the first 5 terms

∴ n = 5

∵
`an=ar^(n-1)`

∵ At n = 1, First term = a

∴ The first term = 2

∵ At n = 2,
`a2=2(-(2)/(3))^(2-1)=2(-(2)/(3))`

∴ The second term =
`-(4)/(3)`

∵ At n = 3,
`a3=2(-(2)/(3))^(3-1)=2(-(2)/(3))^(2)=2((4)/(9))`

∴ The third term =
`(8)/(9)`

∵ At n = 4,
`a4=2(-(2)/(3))^(4-1)=2(-(2)/(3))^(3)=2(-(8)/(27))`

∴ The fourth term =
`-(16)/(27)`

∵ At n = 5,
`a5=2(-(2)/(3))^(5-1)=2(-(2)/(3))^(4)=2((16)/(81))`

∴ The fourth term =
`(32)/(81)`

∴ The first five terms are 2,
`-(4)/(3)` ,
`(8)/(9)` ,
`-(16)/(27)` ,
`(32)/(81)`