Square of sum of four consecutive positive integers is 670 more than t

1.	Square of sum of four consecutive positive integers is 670 more than the sum of square those four consecutive positive integers, then find the average of the four numbers.1. 92. 12.53. 7.54. 6
Answer» Correct Answer - Option 3 : 7.5 GIVEN: Square of sum of four consecutive positive integers is 670 more than the sum of square those four consecutive positive integers. CALCULATION: Let consecutive integers are ‘x’, ‘x + 1’, ‘x + 2’ and ‘x + 3’ Square of sum of four consecutive positive integers = [x + (x + 1) + (x + 2) + (x + 3)]² = (4x + 6)² = 16x² + 36 + 48x Sum of square those four consecutive positive integers = x² + (x + 1)² + (x + 2²) + (x + 3)² = x² + x² + 1 + 2x + x² + 4 + 4x + x² + 9 + 6x = 4x² + 12x + 14 Now, (16x² + 36 + 48x) – (4x² + 12x + 14) = 670 12x² + 36x – 648 = 0 x² + 3x – 54 = 0 x = 6 Required average = [x + (x + 1) + (x + 2) + (x + 3)]/4 = (4x + 6)/4 = x + 1.5 = 6 + 1.5 = 7.5

Square of sum of four consecutive positive integers is 670 more than the sum of square those four consecutive positive integers, then find the average of the four numbers.1. 92. 12.53. 7.54. 6

Answer» Correct Answer - Option 3 : 7.5

GIVEN:

Square of sum of four consecutive positive integers is 670 more than the sum of square those four consecutive positive integers.

CALCULATION:

Let consecutive integers are ‘x’, ‘x + 1’, ‘x + 2’ and ‘x + 3’

Square of sum of four consecutive positive integers = [x + (x + 1) + (x + 2) + (x + 3)]² = (4x + 6)² = 16x² + 36 + 48x

Sum of square those four consecutive positive integers = x² + (x + 1)² + (x + 2²) + (x + 3)² = x² + x² + 1 + 2x + x² + 4 + 4x + x² + 9 + 6x = 4x² + 12x + 14

Now,

(16x² + 36 + 48x) – (4x² + 12x + 14) = 670

12x² + 36x – 648 = 0

x² + 3x – 54 = 0

x = 6

Required average = [x + (x + 1) + (x + 2) + (x + 3)]/4 = (4x + 6)/4 = x + 1.5 = 6 + 1.5

= 7.5

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