1. ## Artihmetic 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

2. Originally Posted by joeblack437
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
the second equation should be $a^3 + 3a^2d + 2ad^2 = 45$

solve for either $a$ or $d$ in the first equation, and plug it into the second and solve, i'd solve for $d$ to begin with

3. 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: . $a,\:a+d,\:a+2d$

Their sum is 15: . $a + (a+d) + (a+2d)\:=\:15\quad\Rightarrow\quad d \:=\:5-a$ .[1]

Their product is 45: . $a(a + d)(a + 2d)\:=\:45$ .[2]

Substitute [1] into [2]: . $a(a + [5-a])(a + 2[5-a]) \:=\:45$

. . This simplifies to: . $a^2 - 10a + 9\:=\:0$

. . which factors: . $(a - 1)(a - 9) \:=\:0$

. . and has roots: . $a \:=\:1,\,9$

Substitute into [1]: . $d \:=\:4,\,-4$

There are two sequences: . $1,\,5,\,9,\,13,\,17,\,\hdots$
. - . . - . . . . . . . . . . . . . . $9,\,5,\,1,\,\text{-}3,\,\text{-}7,\,\hdots$