1,

`a_1=7`

`q=3`

`a_5=a_1*q^4` = `7*3^4` = 567

`a_(12)=a_1*q^(11)` = `7*3^(11)` = 1 240 029

2,

`a_4=a_1*q^3=48`

`a_6=a_1*q^5=192`

`a_6/a_4` = `(cancel(a_1)*q^5)/(cancel(a_1)*q^3)` = `q^2` = `192/48` = 4

I. `q_1` = 2

`a_1=a_4/q^3` = `48/2^3` = 6

II. `q_2=-2`

`a_1=a_4/q^3` = `48/(-2)^3` = -6

3,

`a_1=2048`

`q=1/2`

`a_n=a_1*q^(n-1)` = 0,0625

`2^(11)*(1/2)^(n-1)=1/16`

`(1/2)^(n-1)=2^(-4-11)`

`2^(1-n)=2^(-15)`

1-n = -15

-n=-16

n = 16

4,

`a_1=10`

`q=3`

`S_(10)` = `a_1*(q^(10)-1)/(q-1)` = `10*(3^(10)-1)/(3-1)` = 295 240

5,

`a_1=300`

`q=1.1`

a,

`a_(10)=a_1*q^(10-1)` = `300*1.1^9` = 707,38 m-t aszfaltoztak le a 10. napon.

b,

`S_n=a_1*(q^n-1)/(q-1)` = `300*(1.1^n-1)/(1.1-1)` = `300/0.1*(1.1^n-1)` = 21000

`1.1^n-1=7`

`1.1^n=8`

`n=(lg8)/(lg1.1)` `approx` 21,81

A 22. napon végeznek az aszfaltozással.