Question no. 5, how to find a
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when a,b,c are in AP b-a=c-b hence 2b=a+c
2(3a²+6a+6)=a+2a+2+4a²+5a+4
=>6a²+12a+12=5a²+7a+6
=>a²-5a+6=0
on solving a=3 or 2
2(3a²+6a+6)=a+2a+2+4a²+5a+4
=>6a²+12a+12=5a²+7a+6
=>a²-5a+6=0
on solving a=3 or 2
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