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

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- May 29th 2007, 11:37 AMjoeblack437Artihmetic sequence question
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 - May 29th 2007, 11:55 AMJhevon
- May 29th 2007, 02:29 PMSoroban
Hello, joeblack437!

Quote:

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: .$\displaystyle 1,\,5,\,9,\,13,\,17,\,\hdots$

. - . . - . . . . . . . . . . . . . .$\displaystyle 9,\,5,\,1,\,\text{-}3,\,\text{-}7,\,\hdots$