Correct Answer - 2
From the figure,
`(a)/(x) =(b)/(x+a+b)=(c )/(xa+2b+c)`
`:. ax+a^(2)+ab=bx` (Comparing first two )
or `x=(a^(2)+ab)/(b-a)`
`ax+a^(2)+2ab+ac=cx` (Comparing first and third )
`:. (a^(2)+ab)/(b-a)=(a^(2)+2ab+ac)/(c-a)` ,brgt `a^(2)c-a^(3)+abc-a^(2)b=a^(2)b+2ab^(2)+abc-a^(3)-2a^(2)b-a^(2)c`
or `ac=b^(2)`
Therefore, a,b, and c are in GP.