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The value of x + y + z is 15. If a, x, y...

The value of `x + y + z` is 15. If a, x, y, z, b are in AP while the value of `(1)/(x) + (1)/(y) + (1)/(z) " is " (5)/(3)`. If a, x, y, z b are in HP, then find a and b .

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The correct Answer is:
(i) `a = 1, b = 9`
(i) Now, `a + b = (a + x + y + z + b) - (x + y +z)`
`= (5)/(2) (a +b) - 15` [ since, a, x, y, z are in AP]
`:. " sum " = (5)/(2) (a + b) rArr a + b = 10`..(i)
since, a,x, y, z, b are in HP, then `(1)/(a), (1)/(x), (1)/(y), (1)/(z), (1)/(b)` are in AP
Now, `(1)/(a) + (1)/(b) = ((1)/(a) + (1)/(x) + (1)/(y) + (1)/(z) + (1)/(b)) - ((1)/(x) + (1)/(y) + (1)/(z))`
`= (5)/(2) ((1)/(a)+(1)/(b)) - (5)/(3)`
`rArr (a +b)/(ab) = (10)/(9) rArr ab = (9 xx 10)/(10)` [from Eq. (i)]
On solving Eqs. (i) and (ii), we get
(ii) LHS `= log (x +z) + log (x +z - 2y)`
`= log (x +z) + log [x +z -2 ((2xz)/(x +z))] [ :' y = (2xz)/(x + z)]`
`= log (x +z) + log ((x -z)^(2))/((x +z))`
`= 2 log (x -z) = RHS`
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