Question

The sum of three terms of an A. P. is 24 and the sum of their cubes is 288. Find those three terms
of an A.P
​

Answer 1

Your question should be that The sum of 3 terms of an AP is 12 and sum of their cube is 288.

If it is so then your answer is in the page below.

Answer 2

The three terms of Arithmetic Progression are  8 - $\sqrt{-416}$  , 8  ,   8 +$\sqrt{-416}$  

Step-by-step explanation:

Given as :

For an Arithmetic Progression

The sum of three terms of an A. P = 24

The sum of cube  of three terms of an A. P  = 288

Let The three terms of A.P =  a - d  , a  , a + d

First term = a - d

Second term = a

Third term = a + d

According to question'

Statement I

The sum of three terms of an A. P = 24

i.e ( a - d )  +  a + ( a + d ) = 24

Or, ( a + a + a ) + ( - d + d ) = 24

Or,  3 a + 0 = 24

Or,  3 a = 24

∴      a = $\dfrac{24}{3}$

i.e  a = 8                  ..............1

So, The first term = a = 8

Again

Statement II

The sum of cube  of three terms of an A. P  = 288

i.e  (a - d )³ + a³ + ( a + d )³  = 288

Or, ( a³ - d³ - 3 a² d + 3 a d² ) + a³ + ( a³ + d³ + 3 a² d + 3 a d² ) = 288

Or, ( a³ + a³ + a³ ) + ( - d³ + d³ ) + (  - 3 a² d + 3 a² d ) + ( 3 a d² + 3 a d² ) = 288

Or, 3 a³ + 0 + 0 + 6 a d² = 288

Or, 3 a³ + 6 a d² = 288

Or, a³ + 2 a d² = 96

Now, put the value of a from eq 1

So, 8³ + 2 × 8 d² = 96

Or, 512 + 16 d² = 96

Or,  d² = 96 - 512

Or, d² = - 416

∴   d = $\sqrt{-416}$

So, The three terms are

First term = 8 - $\sqrt{-416}$

Second term = 8

Third term = 8 + $\sqrt{-416}$

Hence, The three terms of Arithmetic Progression are  8 - $\sqrt{-416}$  , 8  ,         8 +$\sqrt{-416}$   Answer