Sum of the terms of Arithmetic Progression (Advanced Problem 1)

Find three numbers in A.P. whose sum is 48 and such that the sum of their squares is 800.

(This problem is from a photocopy of our homework given to us by our professor. But it's the last item that has been solved in our make-up class today)

Find three numbers in A.P. whose sum is 48 and such that the sum of their squares is 800.

First, the detail about the sum gives us the first equation:

a_{1 +} (a_{1 +} d) + (a₁ +2 d) = 48 (from a₁ + a₂ + a₃ = 48)

3a_{1 +} 3d = 48 (Equation 1)

Next, the detail about the product gives us the second equation as:

(a₁)² ₊ (a_{1 +} d) ² + (a₁ +2 d) ² = 800

Simplifying it, we will get

3 a₁^{2 +} 6 a₁d + 5d² = 800 (Equation 2)

Now, we can get the value of a₁ if we express d in terms of a₁using equation 1.

d = 16 - a₁

Substituting the value of d in equation 2, we get


3 a₁² + 6 a₁ (16 - a₁) + 5(16 - a₁)² = 800

Simplifying

a₁² - 32 a₁² + 240 = 0


Getting the roots of the equation, we will get
a₁ = 20                  the other a₁ = 12
d = -4                                     d = 4

Therefore the numbers can be 20, 16, 12
or 12, 16, 20 which are of the same elements.

(I'll elaborate the solutions later on.)

Hello!

Hi, this blog is dedicated to post tutorials about Math and stuff. Thank you! :)