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Question

If a,b,c,d,e,f,g,h are eight consecutive natural numbers, then prove that
a2+d2+f2+g2=b2+c2+e2+h2

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Solution

Solution -
Let numbers be a,a+1,a+2,a+7
Their square sum =a2+(a+1)2+(a+2)2+(a+7)2
=8a2+140+56a
Now LHS
a2+d2+f2+g2=a2+(a+3)2+(a+5)2+(a+6)2
=4a2+70+28a
RHS, b2+c2+e2+h2=(a+1)2+(a+2)2+(a+4)2+(a+7)2
=4a2+70+28a
LHS = RHS

1100699_1188172_ans_abcdacff3d9144b8b64aae7ce651cacf.jpg

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