Sequence Equation

Level pending

If a, b, c are 3 consecutive terms of an Arithmetic Sequence which $a>c$ .

And :

$a+b+c=18$ .

$abc=162$ .

Find $(a-b)(b-c)(c-a)$

1 solution

Ayoub Dergaoui
Jul 20, 2014

$The$ $sequence$ $is$ $written$ $as$ $the$ $forme$ $of$ $U_{n}=U_{p}+(n-p)r$

$a=U_{0}+kr$

$b=U_{0}+(k+1)r$

$c=U_{0}+(k+2)r$

$a+b+c=3U_{0}+3(k+1)r=18$

$So$ $U_{0}=6-(k+1)r$

$abc=(U_{0}+kr)(U_{0}+(k+1)r)(U_{0}+(k+2)r)=162$

$U_{0}+kr=6-r$

$U_{0}+(k+1)r=6$

$U_{0}+(k+2)r=6+r$

$abc=162=6(6-r)(6+r)=162$

$So$ $(6-r)(6+r)=27$

$r=3$ $OR$ $r=-3$

$and$ $a>c$ $so$ $r=-3$

$and$ $a-b=b-c=r=3$

$and$ $c-a=r=-3$

$So$ $(a-b)(b-c)(c-a)=-27$

$The$ $Solution$ $is$ $-3$

I found that a=9,b=6,c=3 and the solution isn't -3

Stefan Alex - 4 years, 11 months ago

c-a=-6 not -3 so the answer must be -3.779.

Archit Agrawal - 4 years, 11 months ago

I agree with Archit. c-a=-6 and not -3

Aditya Dhawan - 4 years, 11 months ago