Question

find four numbers in ap whose sum is 20 and the sum of whose square is 120​

Answer 1

Answer:

Let the terms be a – 3d, a – d, a + d, a+3d

Sum of the terms = (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20

4a = 20

a = 5

Sum of the squares of the term

= (a – 3d)2 + (a – d)2 + (a + d)2 + (a + 3d)2 = 20

a^2 – 6ad + 9d^2 + a^2 – 2ad + d^2 + a^2 + 2ad + d2 + a2 + 6ad + 9d^2 = 120

4a2 + 20d2 = 120 –– (a)

Substituting a = 5 into (a)

4(52) + 20d2 = 120

100 + 20d2 = 120

d = + 1

d = +  1

Thus, the four numbers are:

Taking d = 1

(a – 3d), (a – d), (a + d), (a + 3d)

= (5 – 3), (5 – 1), (5 + 1), (5 + 3)

= 2, 4, 6, 8

Taking d = -1

(a – 3d), (a – d), (a + d), (a + 3d)

= (5 + 3), (5 + 1), (5 - 1), (5 - 3)

= 8, 6, 4, 2

Answer 2

Step-by-step explanation:

= (a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 120

= (a^2 + 9d^2 - 6ad) + (a^2+d^2-2ab) + (a^2+d^2+2ad) + (a^2+9d^2+6ad) = 120

= 4a^2 + 20d^2 = 120

Substitute a = 5 from (2) .

4(5)^2 + 20d^2 = 120

100 + 20d^2 = 120

20d^2 = 20

d = +1 (or) - 1.

Since AP cannot be negative.

Substitute a = 5 and d = 1 in (1), we get

a - 3d, a-d, a+d, a+3d = 2,4,6,8.

Hope this helps!