A wall is in the shape of a trapezium. The first level of the wall is made up of 50 bricks where as the top level has 14 bricks. If the levels differ from each other by 4 bricks, determine the number of; (i)levels of the bricks.
(ii)bricks used to make the wall.

i). 10 levels of the bricks

ii). 320 bricks

Step-by-step explanation:

First level contains number of bricks = 50

Second level will contain = 50 - 4 = 46 bricks

Similarly, 3rd level will contain number of bricks = 46 - 4 = 42

Therefore, sequence formed for the number of bricks in each level of the wall will be,

50, 46, 42........14

This sequence is an arithmetic sequence having,

First term 'a' = 50

Common difference 'd' = 46 - 50 = (-4)

Last term of the sequence = 14

i). Expression representing last term will be,

  Here = nth term

  a = first term

  n = number of term (Number of level of the wall)

  d = common difference

  By substituting these values in the formula,

  14 = 50 + (n - 1)(-4)

  14 - 50 = (-4)(n - 1)

  -36 = -4(n - 1)

  9 = (n - 1)  

  n = 9 + 1

  n = 10

ii). Number of bricks used in the wall = Sum of the sequence

   Expression for the sum of an arithmetic sequence is,

        = 5(100 - 36)

        = 320 bricks

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.