sum of three numbers in AP is 12 and sum of their cubes is 285. Find the
numbers.
Answers
Given :-
sum of three numbers in AP is 12,
Let the three numbers be,
a, a + d, a + 2d
Their sum = 12
a + a + d + a + 2d = 12
3a + 3d = 12
3 ( a + d) = 12
a + d =
a + d = 4
a = 4 - d ---------->1
As per the second condition :-
sum of their cubes is 288 (thanks for the correction @nehaharish)
a³ + (a + d)³ + (a + 2d)³ = 288
Simplifying it we get,
a³ + a³ + d³ + 3a²d + 3ad² + a³ + 8d³ + 6a²d + 12ad² = 288
Bring them together,
a³ + a³ + a³ + d³ + 8d³ + 3a²d + 6a²d + 3ad² + 12ad²
Add all the like terms,
3a³ + 9d³ +9a²d + 15ad² = 288
Substitute the value of 'a' in the above equation,
3 ( 4-d) ³ + 9d³ + 9 (4-d)²d + 15 (4-d) d² = 288
Using (a+b)³ ,we get,
3 (64 - d³ -48d +12d²) + 9d³ + 9 ( 16 + d² - 8d) d +( 60 -15d) d² = 288
192 - 3d³ - 144d + 36d² + 9d³ + 144d + 9d³ - 72d² + 60d² + 15d³ = 288
Multiplying the terms within the brackets,
192 - 3d³ + 9d³ + 9d³ - 15d³ -144d +144d + 36d² -72d² + 60d²
Get the like terms together and perform the required operation,
192 - 18d³ + 18d³ -36d² + 60d² = 288
Subtracting the like terms,
192 + 24d² = 288
Transporting the terms,
24d² = 288 - 192
24d² = 96
d² =
d² = 4
d = √4
d = 2 OR d = -2
Substitute the value of d in equation 1,
Case 1:-
If d = 2
First number,
Second number,
Third number,
Case 2:-
If d = -2
First number,
Second number,
Third number,
•°• The numbers are :-
2,4,6 OR 6,4,2
First condition :-
When d = 2,
Three terms are, 2, 4, 6.
As per the first condition in the question,
•°•a + a + d + a + 2d = 12
Where a = 2, a +d = 4 and a + 2d = 6
2 + 4 + 6 = 12
6 + 6 = 12
12 = 12
LHS = RHS.
Now according to the second condition in the question,
a³ + a+d³ +a+2d³ = 288
2³ + 4³ + 6³ = 288
8 + 64 + 216 = 288
72 + 216 = 288
288 = 288
Hence verified.
Second condition :-
When, d = -2
Three terms are 6,4,2
According to the first statement in the question,
a + a + d + a + 2d = 12
6 + 4 + 2 = 12
10 + 2 = 12
12 = 12
LHS = RHS.
According to the second statement in the question,
a³ + a + d³ + a + 2d ³ = 288
6³ + 4³ + 2³ = 288
216 + 64 + 8 = 288
216 + 72 = 288
288 = 288
LHS = RHS
Hence verified.
Answer:
a
+(a+d)
+(a+2d)
=12
so 3a+3d=12
so a+d=4
Step-by-step explanation:
Also
a^3
+(a+d)^3
+(a+2d)^3
=285
Now if numbers are
3
4
and
5
we get
3+4+5=12
and
3^3+
4^3+
5^3
=27+64+125
=216
So total of cubes being
285 is not possible.