The sum of the first three terms of an arithmetics series is 15, the product of the first 3 terms of the series is 45. find the first 3 terms
i have: 3a+3d = 15 and 3a^2 + 6ad + 2d^2= 45
Hello, joeblack437!
The sum of the first three terms of an arithmetics series is 15.
The product of the first three terms of the series is 45.
Find the first three terms
The first three terms are: .$\displaystyle a,\:a+d,\:a+2d$
Their sum is 15: .$\displaystyle a + (a+d) + (a+2d)\:=\:15\quad\Rightarrow\quad d \:=\:5-a$ .[1]
Their product is 45: .$\displaystyle a(a + d)(a + 2d)\:=\:45$ .[2]
Substitute [1] into [2]: .$\displaystyle a(a + [5-a])(a + 2[5-a]) \:=\:45$
. . This simplifies to: .$\displaystyle a^2 - 10a + 9\:=\:0$
. . which factors: .$\displaystyle (a - 1)(a - 9) \:=\:0$
. . and has roots: .$\displaystyle a \:=\:1,\,9$
Substitute into [1]: .$\displaystyle d \:=\:4,\,-4$
There are two sequences: .$\displaystyle 1,\,5,\,9,\,13,\,17,\,\hdots$
. - . . - . . . . . . . . . . . . . .$\displaystyle 9,\,5,\,1,\,\text{-}3,\,\text{-}7,\,\hdots$