Such as the sum of their square is 800

Let nos be x-d, x, x+d

then (x-d)+(x)+(x+d)=48

=> 3x+d -d =48

3x=48

x=48/3

x=16

now

(x-d)^2 + x^2 + (x+d)^2 = 800

x^2 + d^2 -2xd + x^2 +x^2 + d^2 + 2xd=800

3x^2 + 2d^2 =800

3*(16)^2 + 2d^2 =800

3*256 + 2d^2 =800

768 + 2d^2=800

2d^2=800-768= 32

d^2 = 32/2= 16

d=sqrt(16)

d=4

so numbers are x-d=16-4=12

x=16

x+4=16+4=20

Thanks

Zulfikar Ali

Posted on Jan 27, 2009

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Posted on Jan 02, 2017

The integers of the numbers on three raffle tickets are consecutive integers whose sum is 7,530 are 2509, 2510, and 2511.

Feb 24, 2015 | SoftMath Algebrator - Algebra Homework...

The product of a sum and difference is the difference of two squares. ie (a+b)*(a-b)=a**2 - b**2 [a squared minus b squared]

Jun 03, 2014 | Activision Computers & Internet

That is not clear.

Of positive integers? What is equal to 12? The sum the difference, the product? How many number involved in the operation whose result is 12? Please try again with the appropriate details.

Of positive integers? What is equal to 12? The sum the difference, the product? How many number involved in the operation whose result is 12? Please try again with the appropriate details.

Jan 29, 2014 | Computers & Internet

This sum is equal to 40+48+56+64+72+...+3024+3032+3040+3048= 40+(48+3048)+(56+3040)+(64+3032)+(72+3024)...

There are (3048-40)/8+1=377 terms in sum. (48+3048)=(56+3040)=(64+3032)=(72+3024)=3096 and there are (377-1)/2=188 pairs whose sum is 3096. Only term that is not in such pair is 40.

Finally sum is 40+188*3096=582088.

There are (3048-40)/8+1=377 terms in sum. (48+3048)=(56+3040)=(64+3032)=(72+3024)=3096 and there are (377-1)/2=188 pairs whose sum is 3096. Only term that is not in such pair is 40.

Finally sum is 40+188*3096=582088.

Jul 01, 2011 | Computers & Internet

Compare the terms of the product of the square of binomial and the terms of the product of the sum and difference of two terms. What statements can you make?

Jun 16, 2011 | Computers & Internet

Let X be the lowest of the 3 integers

The sum of the 3 integers can be represented as X + (X+1) + (X+2).

Set this sum equal to 378 and solve for X.

X + (X+1) + (X+2) = 378

3X + 3 = 378

3X = 375

X = 125

Since X is the lowest of the 3 integers the other 2 will be X+1 and X+2 or 126 and 127.

Therefore, the answer is 125, 126 and 127

The sum of the 3 integers can be represented as X + (X+1) + (X+2).

Set this sum equal to 378 and solve for X.

X + (X+1) + (X+2) = 378

3X + 3 = 378

3X = 375

X = 125

Since X is the lowest of the 3 integers the other 2 will be X+1 and X+2 or 126 and 127.

Therefore, the answer is 125, 126 and 127

Aug 23, 2010 | SoftMath Algebrator - Algebra Homework...

The sum of the first counting numbers 1 + 2 + 3 + 4 + 5 +... + is given by

Use , where is the first term in the series, and is the last.

Now to solve this for nos, 1 to 10000 inclusive

Subtract 10000 - 1 = 9999 and add 1 to get 10000 terms in the series.

S = (10000[ 1+10000])/2 = 50,005,000.

This should do it.

Use , where is the first term in the series, and is the last.

Now to solve this for nos, 1 to 10000 inclusive

Subtract 10000 - 1 = 9999 and add 1 to get 10000 terms in the series.

S = (10000[ 1+10000])/2 = 50,005,000.

This should do it.

Jun 05, 2009 | Super Tutor Pre Algebra (ESDPALG)

The two numbers 3 and 24

3 multiplied by 24 =72

24 divided by 3 =8

Enjoy, and thanks for choosing Fixya!

3 multiplied by 24 =72

24 divided by 3 =8

Enjoy, and thanks for choosing Fixya!

Mar 20, 2009 | M2K Garfield: It's All About Math Math...

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