Question

find three consecutive terms in an A.P whose sum is -3 and the product of their cubes is 512 ​

Answer 1

Please see the attachment

Answer 2

The three consecutive numbers are -4,-1,2.

Step-by-step explanation:

Given : Three consecutive terms in an A.P whose sum is -3 and the product of their cubes is 512 .

To find : The consecutive terms ?

Solution :

Let the three consecutive terms in an A.P. are 'a-d,a,a+d'.

The sum of three consecutive terms is -3.

i.e. $a-d+a+a+d=-3$

$3a=-3$

$a=\frac{-3}{3}$

$a=-1$

The product of their cubes is 512.

i.e. $(a-d)^3a^3(a+d)^3=512$

$[(a-d)(a)(a+d)]^3=8^3$

$(a-d)(a)(a+d)=8$

$a(a^2-d^2)=8$

Substitute a=-1,

$(-1)((-1)^2-d^2)=8$

$(-1)(1-d^2)=8$

$-1+d^2=8$

$d^2=9$

$d=\pm3$

The three consecutive terms in A.P is

When a=-1 and d=3,

a-d=-1-3=-4

a=-1

a+d=-1+3=2

When a=-1 and d=-3,

a-d=-1-(-3)=2

a=-1

a+d=-1-3=-4

The three consecutive numbers are -4,-1,2.

#Learn more

Find three consecutive terms in A.P. whose sum is9 and the product of their cubes is 3375​

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