For a, b 0 , let 5 a-b, 2 a+b, a+2 b be in A.P. and (b+1)^2, a b+1,(a-1)^2 are in G.P., then compute (a⁻¹+b⁻¹ )

Step-by-step solution

For the given A.P., we have 2(2 a+b)=(5 a-b)+(a+2 b) b=2 a ( i ) Also for the given G.P., we have (a b+1)^2=(a-1)^2(b+1)^2 ( ii ) Putting b=2 a from (i) in (ii), we get a=0,-2 or 1 4 But a 0 , so a = 1 4 and b =2 a = 1 2 Hence, (a⁻¹+b⁻¹ )=2+4=6