Question

find four numbers in a p whose sum is 20 and sum of whose squares is 120​

Answer 1

$\huge{\boxed{\boxed{\mathfrak{\underline{Solution:}}}}}$

Let the numbers are (a - d), (a + d), (a + 3d) and (a - 3d).

According to 1st condition,

=> (a - d) + (a + d) + (a + 3d) + (a - 3d) = 20

=> 4a = 20

=> a = 5

According to 2nd condition,

[(a - d)² + (a + d)² + (a + 3d)² + (a - 3d)²] = 120

=> [(5 - d)² + (5 + d)² + (5 + 3d)² + (5 - 3d)²] = 120

Separate same terms,

By using identity,

∵ (a - b)² + (a + b)² = 2(a² + b²)

So,

=> 2(25 + d²) + 2(25 + 9d²) = 120

=> 50 + 2d² + 50 + 18d² = 120

=> 20d² + 100 = 120

=> 20d² = 20

=> d² = 1

=> d = ± 1

So, numbers are

=> a + 3d = 2

=> a + d = 4

=> a - d = 6

=> a - 3d = 8.

Answer 2

Explanation:-

Given :-

Sum of number in A.P is 20.

Sum of squares of number in A.P is 140.

Solution:-

Let the four numbers in A.P will be

a + 2d , a + d , a - 2d , a - d.

A/Q

→$a + d + a + 2d + a -2d + a - d = 20$

→$4a = 20$

→$a = \dfrac{20}{4}$

→$a = 5$

→$(a+d)^2 + (a+2d)^2 + (a-d)^2 + (a-2d)^2 = 140$

→$a^2 + d^2 +2ad + a^2 + 4d^2 + 4ad + a^2 +d^2 -2ad + a^2 +4d^2 -4ad = 140$

→$4a^2 + 10 d^2 = 140$

• put the value of a.

→$4 (5)^2 + 10 d^2 = 140$

→$100 +10d^2 = 140$

→$10d^2 = 140 -100$

→$10d^2 = 40$

→$d^2 = 4$

→$d = \pm 2$

hence,

The numbers will be 7 , 3 , 1 , 9.