If a, x, y, z, b are in Arithmetic Progressio | vedclass

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(C) For the $A$.$P$. $a, x, y, z, b$,the sum is $x + y + z = 15$. Since $x, y, z$ are terms between $a$ and $b$,we have $x+z = a+b$ and $y = \frac{a+b

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$9, 1$

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(C) For the $A$.$P$. $a, x, y, z, b$,the sum is $x + y + z = 15$. Since $x, y, z$ are terms between $a$ and $b$,we have $x+z = a+b$ and $y = \frac{a+b}{2}$.Thus,$x+y+z = \frac{3}{2}(a+b) = 15$,which implies $a+b = 10$ (Equation $1$).For the $H$.$P$. $a, x, y, z, b$,the reciprocals $\frac{1}{a}, \frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{b}$ are in $A$.$P$.Similarly,$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{3}{2}(\frac{1}{a} + \frac{1}{b}) = \frac{5}{3}$.This implies $\frac{1}{a} + \frac{1}{b} = \frac{10}{9}$,so $\frac{a+b}{ab} = \frac{10}{9}$.Substituting $a+b=10$,we get $\frac{10}{ab} = \frac{10}{9}$,so $ab = 9$ (Equation $2$).Solving $a+b=10$ and $ab=9$,the quadratic equation $t^2 - 10t + 9 = 0$ gives $(t-9)(t-1) = 0$.Thus,the values are $a, b = 9, 1$.

If $a, x, y, z, b$ are in Arithmetic Progression ($A$.$P$.) such that $x + y + z = 15$,and if $a, x, y, z, b$ are in Harmonic Progression ($H$.$P$.) such that $1/x + 1/y + 1/z = 5/3$,find the values of $a$ and $b$.

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