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three positive numbers are in Arithmetic Progression (A.P). the sum of the squares of the three numbers is 155. while the sum of the numbers is 21. if the common difference is positive. find the values of x and y

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mathmate
Given:
Three numbers in an AP, all positive.
Sum is 21.
Sum of squares is 155.
Common difference is positive.

We do not know what x and y stand for.  Will just solve for the three numbers in the AP.
Let m=middle number, then since sum=21, m=21/3=7
Let d=common difference.
Sum of squares
(7-d)^2+7^2+(7+d)^2=155
Expand left-hand side
3*7^2-2d^2=155
d^2=(155-147)/2=4
d=+2 or -2
=+2  (common difference is positive)

Therefore the three numbers of the AP are
{7-2,7,7+2}, or
{5,7,9}


abidemiokin

The required arithmetic sequence will be 5, 7, and 9 if the sum of the squares of the three numbers is 155. while the sum of the numbers is 21

Let the three positive numbers be a-d, a and a + d

If the sum of the numbers is 21, this means that a-d+a+a+d = 21

3a -d + d = 21

3a = 21

a = 21/3

a = 7

The sequence of numbers will be 7-d, 7 and 7+d

If the squares of the three numbers is 155, hence;

[tex](7-d)^2+7^2+(7+d)^2=155\\49-14d+d^2+49+49+14d+d^2=155\\49++49+49+2d^2=155\\147+2d^2=155\\2d^2=155-147\\2d^2=8\\d^2=4\\d =2[/tex]

Hence the required arithmetic sequence will be 5, 7 and 9

Learn more here: https://brainly.com/question/15021002

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