document.write( "Question 456820: Given three consecutive integers, what is the difference between the average of their squares and the square of their average? \n" ); document.write( "

Algebra.Com's Answer #313540 by spacesurfer(12) $\"\"$ $\"About$

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Did you try to do this your self? Just do the math:\r
\n" ); document.write( "\n" ); document.write( "If a = first integer, then a+1, and a+2 are the other consecutive integers. The average of the squares is:\r
\n" ); document.write( "\n" ); document.write( " $\"%28a%5E2+%2B+%28a%2B1%29%5E2+%2B+%28a%2B2%29%5E2%29%2F3\"$ = $\"a%5E2+%2B+2a+%2B+5%2F3\"$ \r
\n" ); document.write( "\n" ); document.write( "The square of the average is:\r
\n" ); document.write( "\n" ); document.write( " $\"%28%28a%2Ba%2B1%2Ba%2B2%29%2F3%29%5E2\"$ = $\"a%5E2%2B2a%2B1\"$ \r
\n" ); document.write( "\n" ); document.write( "Hence, the difference is 5/3 - 1 = 2/3.\r
\n" ); document.write( "\n" ); document.write( "If the integers didn't increase by 1 and instead increased by b, so like instead of 3,4,5, it was 3, 3+b, 3+2b, for any b (if b = 2, then 3, 5, 7), then the difference is $\"2b%5E2%2F3\"$ .\r
\n" ); document.write( "\n" ); document.write( "if b = 1, then this reduces to 2/3, which we got earlier. \n" ); document.write( "

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