1,

`a_1` = 5

`a_(21)` = `a_1+20*d` = 17

`a_(21)-a_1=20*d` = 17-5 = 12

`d=12/20` = `3/5`

`a_n=a_1+(n-1)*d`

`a_n=5+(n-1)*3/5=8` /-3

`(n-1)*3/5=3` /*(5/3)

n-1 = 5

n = 6

A sorozat 6. tagja a 8.


2,

`a_1=x`

d = 2

`a_(55)=a_1+(55-1)*2` = x + 108

`S_(55)=(a_1+a_n)*n/2` = `(x+x+108)*55/2` = 3905

`(x+54)*55=3905`

x + 54 = 71

x = 17 = `a_1`

`a_(55)` = `a_1+(55-1)*d` = `17+54*2` = 125


3,

`x=2000*^2^(18/3)` = `2000*2^6` = `2000*64` = `color(red)("128 000")`.

4,

I.

`a_1` = 1000

`a_(22)` = `a_1+21*d` = `100+21*100` = 2200

`S_(22)` = `(a_1+a_(22))*22/2` = `(1000+2200)*22/2` = `color(red)("35 200")` Ft-ot kap.


II.

`b_1` = 1000

`q=1.05`

`b_(22)` = `b_1*1.05^21` = 2786

`S_(22)` = `b_1*(q^22-1)/(q-1)` = `1000*(1.05^(22)-1)/(1.05-1)` = `color(green)("38 505")`.

Kezdetben lassabban nő, de ha hatványozottan nő, akkor később gyorsabban fog nőni.



5,

`2000000*(1+x/100)^3=2590058`

`(1+x/100)^3=2590058/2000000` = 1,295

`1+x/100=root(3)(1.295)` = 1,09

`x/100` = 0,09

x = `color(red)("9 %")` a kamat.