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Explanation

Let `b/a = c/b =r`
`:." "b=ar, c=ar^3`
Given, `(a+b+c)/3 = b+2 rArr 1+b/a +c/a =3 (b/a)+6/a`
`rArr" "1+r+r^2 =3r+6/a`
Now, for a = 6, only we get r = 0, 2 [ rationa ]
So, r=2
`rArr" "(a,b,c)=(6, 12, 24)`
`:." "(a^2+a-14)/(a+1)=(36+6-14)/(6+1) =4`