Let, a,ar,ar2→G.P.
It is given that sum of any two sides is greater than the third side.
⇒a+ar>ar2,a+ar2>ar,ar+ar2>a
Case-I: r2−r−1<0
⇒r=21±1−4×1×(−1)
⇒r=21±5
⇒r∈(21−5,21+5)...(i)
Case-II: r2−r+1>0 is always true.
Case-III: r2+r−1>0
⇒r∈(−∞,2−1−5)∪(2−1+5,∞)...(ii)
Taking intersection of (i)and(ii)
⇒r∈(2−1+5,21+5)
As r>1
⇒r∈(1,21+5)
⇒[r]=1and[−r]=−2
⇒3[r]+[−r]=3−2
⇒3[r]+[−r]=1