Given that `x+y+z =15` when a,x,y,z,b are in A.P and `1/x+1/y+1/z=5/3 – InterviewSolution

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1.	Given that `x+y+z =15` when a,x,y,z,b are in A.P and `1/x+1/y+1/z=5/3 when a,x,y,z,b are in H.P .ThenA. G.M of a and b is 3B. one possible value of a + 2b is 11C. A.M of a and b is 6D. greatest value of a-b is 8
Answer» Correct Answer - A::B::D `x+y+z=3((a+b)/2)` `rArr15-3((a+b))/2` `rArra+b=10` `1/x+1/y+1/z=(3(1/a+1/b))/2` `rArr5/3=(3(a+b))/(2ab)=(3xx10)/(2ab)` `rArrab=9` (2) From (1) and (2), a=9,b=1 or a=1 and b=9. Hence, G.M. `=sqrt(ab)=3,a+2b=11` or 19.

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