iu) Find three conseCutive terms in A.P. whose sum is 9 and the product of their cubes is 3375.
Answers
Answer:
1, 3 and 5.
Step-by-step explanation:
a + (a + d) +( a + 2d) = 9 where a is the first of the numbers.
3a + 3d = 9
a + d = 3
d = 3 - a
a^3 * (a + d)^3 * ( a + 2d)^3 = 3375
Substituting for d:-
a^3 * (a +3 - a)^3 * (a+ 6 - 2a)^3 = 3375
a^3(3)^3(6-a)^3 = 3375
27a^3(6-a)^3 = 3375
The second number is cube root of 27 = 3.
a^3(6-a)^3 = 3375/27
a^3(6-a)^3 = 125
5^3 = 125.
so a^3 = 1 and (6 - 1)^3 = 125
The three numbers are 1, 3 and 5.
Answer: 1,3,5
Step-by-step explanation:
a + (a + d) +( a + 2d) = 9 where a is the first of the numbers.
3a + 3d = 9
a + d = 3
d = 3 - a
a^3 * (a + d)^3 * ( a + 2d)^3 = 3375
Substituting for d:-
a^3 * (a +3 - a)^3 * (a+ 6 - 2a)^3 = 3375
a^3(3)^3(6-a)^3 = 3375
27a^3(6-a)^3 = 3375
The second number is cube root of 27 = 3.
a^3(6-a)^3 = 125
5^3 = 125.
so a^3 = 1 and (6 - 1)^3 = 125