Question

If sum of three terms in an AP is 12 and sum of their cubes is 288, what are these terms

Answer 1

Answer:

Three terms are 2, 4 and 6

Step-by-step explanation:

Let three terms of AP are (a - d ), a and (a + d)

Given that sum of three terms of AP = 12

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

3a = 12

a = 4

Also, sum of cubes of three terms of AP = 288

$(\,4 - d)\,^3 + (\,4)\,^3 + (\,4 + d)\,^3 = 288\\\\ (\,64 - 3(\,16)\,(\,d)\, + 3(\,d)\,^2(\,4)\, - (\,d)\,^3) + [\, 64]\, + [\,64 + 3(\,16),(\,d)\, + 3(\,d)\,^2(\,4)\, +(\,d)\,^3]\, = 288\\\\ 64 - 48d + 12d^2 + d^3 + 64 + 64 + 48d + 12d^2 + d^3=288\\\\ 192 + 24d^2 = 288\\\\ 24d^2 = 288 - 192\\\\ 24d^2 = 94\\\\ d^2 = 4\\\\ d = \pm 2$

When d = 2

Three numbers are, 2, 4, 6

When d = -2

Three numbers are 6, 4, 2

Answer 2

Answer:

2,4,6 or 6,4,2

Step-by-step explanation:

let the  terms be a-d, a ,a+d

according to the given condition,

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

thus, 3a = 12

therefore, a = 4... [1]

according to second condition

(a-d)^3 + a^3 + (a+d)^3 = 288

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

3a^3 + 6 ad^2=288

but a= 4

therefore, 3(4)^3 + 6 ( 4) ( d)^2 = 288

192 + 24d^2 = 288

• 24d^2 = 96

d^2 = 4

d = +-(2)

therefore the terms are,

if a=4 and d=2

t1 = a-d = 2

t2 = a = 4

t3 = a+d = 6

OR

if a=4 and d=-2

t1 = a-d = 6

t2 = a= 4

t3 = a+d = 4