Question

The sum of three numbers in AP is 33,and the sum of their squares is 461.Find the numbers

Answer 1

Given:

The sum of three numbers in A.P is 33, and the sum of the squares is 461.

To Find:

The three numbers are?

Solution:

The given problem can be solved using the concepts of Arithmetic Progression.

1. Let the three terms be a-d, (a), a+d.

=> a-d + a + a +d = 33,

=> 3a = 33,

=> a = 11.

2. The sum of the squares is 461,

=> (a-d)² + a² + (a+d)² = 461,

=> a² + d² -2ad + a² + a² + d² + 2ad = 461,

=> 3a² + 2d² = 461,

=> 3x121 + 2d² = 461,

=> 363 + 2d² = 461,

=> 2d² = 461-363,

=> 2d² = 98,

=> d² = 49,

=> d = +7 (OR) d = -7.

3. The terms of the A.P are 11 - 7, 11, 11 + 7 (OR) 11-(-7), 11, 11 + 7,

=> The terms of the A.P are 4, 11, 18 (OR) 18, 11, 4.

Therefore, the numbers are 4, 11, 18 (OR) 18, 11, 4.

Answer 2

Given :

The sum of three numbers in AP is 33, and the sum of their squares is 461

To Find :

The numbers

Solution :

Let a-d, a, a+d be the 3 numbers in AP, where d is the common difference.

Sum of the three numbers = 33

So, (a-d) + a + (a + d) = 33

⇒                         3a  = 33

∴                            a  =  11

Given, Sum of their squares               = 461

∴  (a-d)² + a² + (a+d)²                           = 461

⇒ a² - 2ad + d² + a² + a² + 2ad + d²   = 461

⇒               3a² + 2d²                            = 461

⇒                         2d²                           =  461 - 3a²

⇒                         2d²                           =  461 - 3(11)²

⇒                         2d²                           =  461 - 363

⇒                         2d²                           =  98

⇒                           d                            = $\sqrt{49}$

∴                            d                            = ± 7

Taking the positive value of d we have,

(a-d) = 11 - 7 = 4

   a   =  11

(a+d) = 11 + 7 = 18

Hence, the numbers are 4, 11 and 18