# The Sum of Three Numbers in A.P. is 12 and Sum of Their Cubes is 288. Find the Numbers. - Mathematics

The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers.

#### Solution 1

Let the three numbers be a d, a and a + d.

S3 = 12

a d + a + a + d = 12

⇒ 3a = 12

a = 4          .....(1)

(4d)3+43+(4+d)3=288

64+d3+48d+12d2+64+64d348d+12d2=288

24d2=96

d=±2

If d = 2, then the numbers are (4 − 2), 4 and (4 + 2), that is, 2, 4 and 6.

If d = −2, then the numbers are (4 + 2), 4 and (4 − 2), that is, 6, 4 and 2

#### Solution 2

In the given problem, the sum of three terms of an A.P is 12 and the sum of their cubes is 288.

We need to find the three terms.

Here,

Let the three terms b (a - d), a, (a + d) where a is the first term and d is the common difference of the A.P

So,

(a - d) + a + (a + d) = 12

3a = 12

a= 12/3

a = 4

Also, it is given that

(a - d)^3 + a^3 + (a + d)^3  = 288

So, using the properties:

(a - b)^3 = a^3 - b^3 + 3ab^2  - 3a^2b

(a + b)^3 = a^3 + b^3 + 3ab^2 + 3a^2b

We get

(a - d)^3 + a^3 + (a + d)^3  = 288

a^3 - d^3 - 3a^3d^2a + a^3 + a^3 + d^3 + 3a^2d + 3d^2a = 288

3a^3 + 6d^2a = 288

a^3 + 2d^2 a = 96

Further solving for by substituting the value of a, we get,

(4)^3 + 2d^2 (4) = 96

64 + 8d^2 = 96

8d^2 = 96 - 64

d^2 = 32/8

On further simplification we get

d = 4

d= sqrt4

d = +2`

Now, here d can have two values +2 and -2

So, on substituting the values of a = 4 and d = -2 in three terms we get

First term = a - d

So,

a - d = 4 - 2

= 2

Second term = a

So,

a = 4

Third term = a + d

So,

a + d = 4  + 2

= 6

Also on substituting the values of a = 4 and d = -2 in three terms, we get

First term = a - d

So,

a - d  = 4 - (-2)

= 4 + 2

= 6

Second term = a

So,

a = 4

Third term = a + d

So,

a + d = 4 + (-2)

= 4 - 2

= 2

Therefore the three terms are 2, 4 and 6 or 6, 4 and 2

Concept: Arithmetic Progression
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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.5 | Q 7 | Page 30