Question

Find the term of terms in a gp​

Answer 1

There are 7 terms in the GP

Explanation:

In the geometric progression, there is a constant ratio between each two consecutive terms

The rule of the nth term of the geometric progression is

`a(n)=a(r)^(n-1)`, where

• a is the first term

• r is the constant ratio

• n is the position of the term in the sequence

∵ The first term = 5
`(1)/(3)`

∴ a = 5
`(1)/(3)`

∵ The last term =
`(243)/(256)`

∴ a(n) =
`(243)/(256)`

∵ The common ratio =
`(3)/(4)`

→ Substitute these values in the rule above to find n

∵
`(243)/(256)` = 5
`(1)/(3)` .
`[(3)/(4)]^(n-1)`

→ Divide both sides by 5
`(1)/(3)`

∴
`(729)/(4096)` =
`[(3)/(4)]^(n-1)`

→ Let us find how many 3 in 729

∵ 729 ÷ 3 = 243 ÷ 3 = 81 ÷ 3 = 27 ÷ 3 = 9 ÷ 3 = 3 ÷ 3 = 1

∴ There are 6 three in 729

∴ 729 =
`3^(6)`

→ Let us find how many 4 in 4096

∵ 4096 ÷ 4 = 1024 ÷ 4 = 256 ÷ 4 = 64 ÷ 4 = 16 ÷ 4 = 4 ÷ 4 = 1

∴ There are 6 four in 4096

∴ 4096 =
`4^(6)`

∵
`(729)/(4096)` =
`(3^(6) )/(4^(6))` =
`[(3)/(4)]^(6)`

∴
`[(3)/(4)]^(6)` =
`[(3)/(4)]^(n-1)`

∵ The bases are equal

∴ Their exponents are equal

∴ 6 = n - 1

→ Add 1 to both sides

∴ 6 + 1 = n - 1 + 1

∴ 7 = n

∴ There are 7 terms in the GP