If 9, x, y, z, a are in A.P. such that x + y | vedclass

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(A) Case $1$: If $9, x, y, z, a$ are in $A.P.$,then the sum of the terms is $9 + x + y + z + a = \frac{5}{2}(9 + a)$.Given $x + y + z = 15$,we have $9

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(A) Case $1$: If $9, x, y, z, a$ are in $A.P.$,then the sum of the terms is $9 + x + y + z + a = \frac{5}{2}(9 + a)$.Given $x + y + z = 15$,we have $9 + 15 + a = \frac{5}{2}(9 + a)$.$24 + a = \frac{45 + 5a}{2}$.$48 + 2a = 45 + 5a$.$3a = 3 \Rightarrow a = 1$.Case $2$: If $9, x, y, z, a$ are in $H.P.$,then $\frac{1}{9}, \frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{a}$ are in $A.P.$The sum of these terms is $\frac{1}{9} + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{a} = \frac{5}{2}\left(\frac{1}{9} + \frac{1}{a}\right)$.Given $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3}$,we have $\frac{1}{9} + \frac{5}{3} + \frac{1}{a} = \frac{5}{2}\left(\frac{1}{9} + \frac{1}{a}\right)$.$\frac{16}{9} + \frac{1}{a} = \frac{5}{18} + \frac{5}{2a}$.$\frac{1}{a} - \frac{5}{2a} = \frac{5}{18} - \frac{16}{9}$.$-\frac{3}{2a} = \frac{5 - 32}{18} = -\frac{27}{18} = -\frac{3}{2}$.$-\frac{3}{2a} = -\frac{3}{2} \Rightarrow a = 1$.

If $9, x, y, z, a$ are in $A.P.$ such that $x + y + z = 15$,and $9, x, y, z, a$ are in $H.P.$ such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3}$,then the value of $a$ is:

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