Question

Find three consecutive terms in an A.P. who's sums is -3 and the product of their cubes is 512

Answer 1

For three consecutive no.s in AP

b=(a+c)/2-----------(1)

Given

a+b+c=-3

On substituting (1)

a+(a+c)/2+c=-3

(2a+a+c+2c)/2=-3

3a+3c=-6

On dividing both sides by 3

a+c=-2-------------(i)

Given :-

$a^{3} b^{3}c^{3}$=512

Cube rooting both sides

abc=8

On substituting (1)

a((a+c)/2)c=8

on substituting (i)

a(-2/2)c=8

-ac=8

so therefore

c=8/-a

therefore substituting in (i)

a+c=-2

a+8/-a=-2

(-a^2+8)/-a=-2

-a^2+8=2a

0=a^2+2a-8

0=a^2 +4a-2a-8

0=a(a+4)-2(a+4)

0=(a-2)(a+4)

therefore a =2 or -4

when a=2

c=-2-2      [from (i)]

c=-4

b=(a+c)/2

 = (-4+2)/2

 = -2/2

b  = - 1

when a=-4

c=4-2     [from (i)]

c=2

b=(a+c)/2

 =(-4+2)/2

 = -2/2

b =-1