we have 10 numbers ±1, ±2, ±3, ±4, ±5 Let S is the sum of all possible products taken two at a time. We know that, for two numbers (a+b)2=(a2+b2)+2ab=(a2+b2)+2S For three numbers (a+b+c)2=(a2+b2+c2)+2S So, (1+2+3+......−5)2=(12+22+....+(−5)2)+2S 0=2(1+4+9+16+25)+2S S=−(1+4+9+16+25)=−55