Find four numbers in A.P whose sum is 40 and the sum of whose squares is 480
Answers
Step-by-step explanation:
Let the four numbers be (a + 3d), (a + d), (a – 3d), (a – d).
Sum of the four numbers is 40.
a + 3d + a + d + a – 3d + a – d = 40
4a = 40
a = 10
Sum of their squares is 480.
(a + 3d)^2 + (a + d)^2 + (a – 3d)^2 + (a – d)^2 = 480
4a^2 + 20d^2 = 480
10^2 + 5d^2 = 120
100 + 5d^2 = 120
5d^2 = 20
d^2 = 4
d = 2, -2
Hence the four numbers are 4, 8, 12, and 16.
Given:
Sum of numbers in an AP = 40
Sum of its squares = 480
To Find:
The four numbers
Solution:
Let the first number be - (a + 3d),
Second - (a + d),
Third - (a – 3d),
Fourth - (a – d).
Now, since the sum of the four numbers is 40.
Therefore,
a + 3d + a + d + a – 3d + a – d = 40
4a = 40
a = 10
Sum of their squares is 480.
Therefore,
(a + 3d)² + (a + d)² + (a – 3d)² + (a – d)²= 480
4a² + 20d²= 480
10² + 5d² = 120 ( Putting value of a = 10 )
100 + 5d²= 120
5d²= 20
d² = 4
d = 2, -2
Taking , d = +2
(a - 3d) = 10 - 3 × 2 = 4.
(a - d) = 10 - 2 = 8.
(a + d) = 10 + 2 = 12.
(a + 3d) = 10 + 3*2 = 16.
Taking, d = (-2).
(a - 3d) = 10 - 3 × (-2) = 16.
(a - d) = 10 - (-2) = 12.
(a + d) = 10 - 2 = 8.
(a + 3d) = 10 + 3*(-2) = 4.
Answer: The required four numbers will be (4,8,12,16) or, (16,12,8,4).