Question

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

Answer 1

Let the terms be, a, a+d, a+2d. Given that sum is 12.

That is, a+(a+d)+(a+2d) = 12
That is, 3a+3d = 12,
Meaning, a+d=4

Since sum is 12, numbers can be (1,4,7) or (2,4,6) or (3,4,5).

But it's given that sum of cubes is 408. This is satisfied only by (1,4,7).

Hence the numbers are 1,4,7.

Answer 2

Answer:

the three numbers of the series are 1, 4, 7

Step-by-step explanation:

let the numbers be (a-d), a, (a+d)

given sum of the numbers  = 12

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

hence, 3a  = 12 and a = 4

and sum of their cubes = 408

⇒ (a-d)³+ a³+ (a+d)³ = 408

$(a+b)^{3} = a^{3}+b^{3}+3ab^{2} +3ba^{2} \\(a-b)^{3} = a^{3}-b^{3}+3ab^{2} -3ba^{2}$

⇒a³ -b³ -3a²d+ 3ad² +a³ + a³ +b³ + 3a²d + 3ad² = 408

⇒ 3a³+6ad² = 408

substitute value for 'a'

⇒3×4³ + 6×4×d² = 408

⇒ 192 + 24d² = 408

d = $\sqrt{\frac{408-192}{24} }$ = √9

d = ±3

when d is 3, the numbers are 1, 4, 7

when d is -3, the numbers are 7, 4, 1

hence, the three numbers of the series are 1, 4, 7