Question

sum of three numbers in AP is 12 and sum of their cubes is 285. Find the
numbers.​

Answer 1

$\bf{\huge{\boxed{\underline{\rm{\blue{Answer:}}}}}}$

Given :-

sum of three numbers in AP is 12,

Let the three numbers be,

a, a + d, a + 2d

Their sum = 12

a + a + d + a + 2d = 12

3a + 3d = 12

3 ( a + d) = 12

a + d = $\bf\large\frac{12}{3}$

a + d = 4

a = 4 - d ---------->1

As per the second condition :-

sum of their cubes is 288 (thanks for the correction @nehaharish)

a³ + (a + d)³ + (a + 2d)³ = 288

Simplifying it we get,

a³ + a³ + d³ + 3a²d + 3ad² + a³ + 8d³ + 6a²d + 12ad² = 288

Bring them together,

a³ + a³ + a³ + d³ + 8d³ + 3a²d + 6a²d + 3ad² + 12ad²

Add all the like terms,

3a³ + 9d³ +9a²d + 15ad² = 288

Substitute the value of 'a' in the above equation,

3 ( 4-d) ³ + 9d³ + 9 (4-d)²d + 15 (4-d) d² = 288

Using (a+b)³ ,we get,

3 (64 - d³ -48d +12d²) + 9d³ + 9 ( 16 + d² - 8d) d +( 60 -15d) d² = 288

192 - 3d³ - 144d + 36d² + 9d³ + 144d + 9d³ - 72d² + 60d² + 15d³ = 288

Multiplying the terms within the brackets,

192 - 3d³ + 9d³ + 9d³ - 15d³ -144d +144d + 36d² -72d² + 60d²

Get the like terms together and perform the required operation,

192 - 18d³ + 18d³ -36d² + 60d² = 288

Subtracting the like terms,

192 + 24d² = 288

Transporting the terms,

24d² = 288 - 192

24d² = 96

d² = $\bf\large\frac{96}{24}$

d² = 4

d = √4

d = 2 OR d = -2

Substitute the value of d in equation 1,

Case 1:-

If d = 2

First number,

$\bf{\large{\boxed{\underline{\rm{\pink{a = 4 - d = 4 - 2 = 2 }}}}}}$

Second number,

$\bf{\large{\boxed{\underline{\rm{\red{a + d = 2 + 2 = 4}}}}}}$

Third number,

$\bf{\large{\boxed{\underline{\rm{\blue{a + 2d = 2 + 2×2 = 2 + 4 = 6}}}}}}$

Case 2:-

If d = -2

First number,

$\bf{\large{\boxed{\underline{\rm{\red{a = 4 - d = 4 - (-2) = 4 + 2 = 6}}}}}}$

Second number,

$\bf{\large{\boxed{\underline{\rm{\green{a + d = 6 + (-2) = 6 - 2 = 4</p><p>}}}}}}$

Third number,

$\bf{\large{\boxed{\underline{\rm{\pink{a + 2d = 6 + 2 (-2) = 6 + (-4) = 2</p><p>}}}}}}$

•°• The numbers are :-

2,4,6 OR 6,4,2

$\bf{\huge{\boxed{\underline{\rm{\blue{Verification:}}}}}}$

First condition :-

When d = 2,

Three terms are, 2, 4, 6.

As per the first condition in the question,

$\bf{\large{\boxed{\underline{\rm{\pink{Sum \: of\: three\: numbers\: in \:AP\: is\: 12,}}}}}}$

•°•a + a + d + a + 2d = 12

Where a = 2, a +d = 4 and a + 2d = 6

2 + 4 + 6 = 12

6 + 6 = 12

12 = 12

LHS = RHS.

Now according to the second condition in the question,

$\bf{\large{\boxed{\underline{\rm{\red{sum\: of\: their\: cubes\: is\: 288,}}}}}}$

a³ + a+d³ +a+2d³ = 288

2³ + 4³ + 6³ = 288

8 + 64 + 216 = 288

72 + 216 = 288

288 = 288

Hence verified.

Second condition :-

When, d = -2

Three terms are 6,4,2

According to the first statement in the question,

$\bf{\large{\boxed{\underline{\rm{\green{Sum \: of\: three\: numbers\: in \:AP\: is\: 12,}}}}}}$

a + a + d + a + 2d = 12

6 + 4 + 2 = 12

10 + 2 = 12

12 = 12

LHS = RHS.

According to the second statement in the question,

$\bf{\large{\boxed{\underline{\rm{\pink{sum \:of \:their \:cubes\: is \:288}}}}}}$

a³ + a + d³ + a + 2d ³ = 288

6³ + 4³ + 2³ = 288

216 + 64 + 8 = 288

216 + 72 = 288

288 = 288

LHS = RHS

Hence verified.

Answer 2

Answer:

a

+(a+d)

+(a+2d)

=12

so 3a+3d=12

so a+d=4

Step-by-step explanation:

Also

a^3

+(a+d)^3

+(a+2d)^3

=285

Now if numbers are

3

4

and

5

we get

3+4+5=12

and

3^3+

4^3+

5^3

=27+64+125

=216

So total of cubes being

285 is not possible.