the sum of 3 terms in AP is 12 and sum of the cubes is 288 find the numbers
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three terms in AP= (a-d),a,(a+d)........(i)
sum of terms a-d+a+a+d= 12
3a= 12
a= 4
sum of cubes
(a-d)^3+a^3+(a+d)^3=288
a^3-d^3-3ad(a-d)+a^3+a^3+d^3+3ad(a+d)
=288
a^3+3ad^2+a^3+3ad^2+a^3=288
3a^3+6ad^2=288
3*4^3+6*4*d^2=288
192+24d^2=288
24d^2=96
d^2=96/24=4
d=√4=2
put in (i)
a-d=4-2= 2
a=4
a+d = 4+2 = 6
three numbers are 2,4,6
sum of terms a-d+a+a+d= 12
3a= 12
a= 4
sum of cubes
(a-d)^3+a^3+(a+d)^3=288
a^3-d^3-3ad(a-d)+a^3+a^3+d^3+3ad(a+d)
=288
a^3+3ad^2+a^3+3ad^2+a^3=288
3a^3+6ad^2=288
3*4^3+6*4*d^2=288
192+24d^2=288
24d^2=96
d^2=96/24=4
d=√4=2
put in (i)
a-d=4-2= 2
a=4
a+d = 4+2 = 6
three numbers are 2,4,6
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