Q2 of 95 Page 68

In the multiplication table made earlier, take a square of nine numbers, instead of four, and mark the numbers at the four corners:


Explain using algebra, why this difference is the same for all such squares.

we can write a row of multiplication table as

x 2x 3x 4x 5x 6x 7x 8x 9x


Next row will be:


(x+1) 2(x+1) 3(x+1) 4(x+1) 5(x+1) 6(x+1) 7(x+1) 8(x+1) 9(x+1)


If we take a general number from the first row as yx then the next number in this row will be (y+1)x.


So, we can obtain a general square of 9 terms from this multiplication table as:



Diagonal sums: (yx)+(y+2)(x+2) = yx+y(x+2)+2(x+2) = yx + yx + 2y +2x +4


= 2yx +2y +2x +4


Y(x+2)+ (y+2)x = (yx +2y)+(yx + 2x) = 2yx +2y +2x


Clearly, the difference between diagonal sums is equal to 4.


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In the multiplication table made earlier, take a square of nine numbers, instead of four, and mark the numbers at the four corners:


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In the multiplication table made earlier, take a square of nine numbers, instead of four, and mark the numbers at the four corners:


Explain using algebra, why this difference is the same for all such squares.

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In the multiplication table made earlier, take a square of nine numbers, instead of four, and mark the numbers at the four corners:


What if we take a square of sixteen numbers?

 2

In the multiplication table made earlier, take a square of nine numbers, instead of four, and mark the numbers at the four corners:


What if we take a square of sixteen numbers?