the sum of the three numbers in AP is 12 and sum of their cubes is 288 find the numbers
Answers
Step-by-step explanation:
2, 4 and 6.
Step-by-step explanation:
Let the three numbers in AP be a,a+d,a+2d
Now we are given that the sum of three numbers in AP is 12
So, a+a+d+a+2d =12a+a+d+a+2d=12
3a+3d =123a+3d=12
a+d =4a+d=4
a =4-da=4−d --1
Now we are given that sum of their cubes is 288.
(a)^3 + (a + d)^3 + (a + 2d)^3 = 288(a)
3
+(a+d)
3
+(a+2d)
3
=288
a^3 + a^3 +d^3 +3a^2d +3ad^2 +a^3 + 8d^3 + 6a^2d +12ad^2 =288a
3
+a
3
+d
3
+3a
2
d+3ad
2
+a
3
+8d
3
+6a
2
d+12ad
2
=288
3a^3 + 9d^3 +9a^2d +15ad^2 =2883a
3
+9d
3
+9a
2
d+15ad
2
=288
3(4-d)^3 + 9d^3 +9(4-d)^2d +15(4-d)d^2 = 2883(4−d)
3
+9d
3
+9(4−d)
2
d+15(4−d)d
2
=288 [using equation 1]
3(64 -d^3 -48d +12d^2 ) + 9d^3 + 9(16 + d^2 -8d) d + (60 -15d)d^2 = 2883(64−d
3
−48d+12d
2
)+9d
3
+9(16+d
2
−8d)d+(60−15d)d
2
=288
192 - 3d^3 - 144d + 36d^2 +9d^3 + 144d + 9d^3 - 72d^2 +60d^2 - 15d^3 = 288192−3d
3
−144d+36d
2
+9d
3
+144d+9d
3
−72d
2
+60d
2
−15d
3
=288
24d^2 = 288 -19224d
2
=288−192
24d^2 =9624d
2
=96
d^2 = \frac{96}{24}d
2
=
24
96
d^2 = 4d
2
=4
d = \pm 2d=±2
For d = 2
a = 4 – d
= 4 – 2
= 2
So,The numbers will be 2, 4 and 6.
For d = - 2
a = 4 - (-2)
= 4 + 2
= 6
The numbers will be 6, 4 and 2.
Hence, the required numbers are 2, 4 and 6.
Answer:
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