Question

The sum of three numbers in A.P is 12 and sum of their cubes is 408. Find the numbers

Answer 1

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

Given :-

• The sum of three numbers in A.P is 12
• Sum of their cubes is 408.

To find :-

• The numbers

Solution :-

Let the three terms in an AP be :

• a - d
• a
• a + d

As per the first condition,

• The sum of three numbers in A.P is 12

$\bf\implies$a - d + a + a + d = 12

$\bf\implies$ 3a = 12

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

a = 4 --->1

As per the second condition,

• sum of their cubes is 408.

(a - d)³ + (a)³ + ( a + d) ³ = 408

Use the identity (a+b)³

a³ - 3a²d + 3ad² - d³ + a³ + a³ + 3a²d + 3ad² + d³ = 408

Bring the like terms together,

a³ + a³ + a³ -3a² d + 3a²d + 3ad² + 3ad² - d³ + d³ = 408

3a³ + 6ad² = 408

3 (4)³ + 6 (4)(d²) = 408

3 × 16 × 4 + 24d² = 408

3 × 64 + 24d² = 408

192 + 24d² = 408

24d² = 408 - 192

24d² = 216

d² = $\bf\large\frac{216}{24}$

d² = 9

d = √9

d = + 3 OR d = - 3

Let's find the numbers by substituting the values of d,

When d = + 3

First term = a - d = 4 - 3 = 1

Second term = a = 4

Third term = a + d = 4 + 3 = 7

When d = + 3

The three terms are : 1,4,7

When d = - 3

First term = a - d = 4 - (-3) = 4 + 3 = 7

Second term = a = 4

Third term = a + d = 4 + (-3) = 4 - 3 = 1

When, d = - 3

The three terms are : 7,4,1

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

For first case :-

• The sum of three numbers in A.P is 12

When d = + 3,

First term = 1 = a - d

Second term = 4 = a

Third term = 7 = a + d

a - d + a + a + d = 12

1 + 4 + 7 = 12

5 + 7 = 12

12 = 12

LHS = RHS.

When, d = - 3

First term = 7 = a - d

Second term = a = 4

Third term = a + d = 1

a - d + a + a + d = 12

7 + 4 + 1 = 12

11 + 1 = 12

12 = 12

LHS = RHS.

For second case :-

• sum of their cubes is 408.

When d = + 3

(a - d) ³ + (a) ³ + ( a + d) ³ = 408

1³ + 4³ + 7³ = 408

1 + 64 + 343 = 408

65 + 343 = 408

408 = 408

LHS = RHS

When, d = - 3,

(a - d) ³ + (a) ³ + (a - d) ³ = 408

7³ + 4³ + 1³ = 408

343 + 64 + 1 = 408

343 + 65 = 408

408 = 408

LHS = RHS

Hence verified.

Answer 2

Step-by-step explanation:

Answer:-

The three number will be

7,4,1

Because

$7 + 4 + 1 = 12$

Also,

Sum of their cube

That is

${7}^{3} + {4}^{3} + {1}^{3 } = 408 \\ 343 + 64 + 1 = 408$