Question

If the nth terms of the progression 5, 10, 20, ... and the progression 1280, 640, 320, ... are equal, then find the value of n.

Answer 1

a = 5

r = 2

`a_n = ar^(n - 1)`

`a_n = 5(2)^(n - 1)`

a = 1280

r = `640/1280`

= `1/2`

`a_n = ar^(n - 1)`

`a_n = 1280(1/2)^(n - 1)`

`5(2)^(n - 1) = 1280(1/2)^(n - 1)`

`(2)^(n - 1) = 1280/5(1/2)^(n - 1)`

`(2)^(n - 1) = 256(1/2)^(n - 1)`

`(2)^(n - 1) = (2^8(1)^(n - 1))/(2)^(n - 1)`

`(2)^(n - 1) = 2^(8 - n - 1)`

`(2)^(n - 1) = (2)^(8 - n + 1)`

`(2)^(n - 1) = (2)^(9 - n)`

n − 1 = 9 − n

n + n = 9 + 1

2n = 10

n = `10/2`

n = 5