x + y + z = 15 if 9,;x,;y,;z,;a are in A.P.; | vedclass

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Question

(a) $x + y + z = 15$, if $9,x,y,z,a$ are in $A.P.$

Sum $ = 9 + 15 + a = \frac{5}{2}(9 + a)$

$ \Rightarrow $$24 + a = \frac{5}{2}(9 + a)$

$ \R

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$1$

Vedclass · Answer

(a) $x + y + z = 15$, if $9,x,y,z,a$ are in $A.P.$

Sum $ = 9 + 15 + a = \frac{5}{2}(9 + a)$

$ \Rightarrow $$24 + a = \frac{5}{2}(9 + a)$

$ \Rightarrow $$48 + 2a = 45 + 5a$

$ \Rightarrow $$3a = 3$$ \Rightarrow $$a = 1$&hellip;..$(i)$

and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3}$, if $9,\;x,\;y,\;z,\;a$ are in $H.P.$

Sum =$\frac{1}{9} + \frac{5}{3} + \frac{1}{a} = \frac{5}{2}\left[ {\frac{1}{9} + \frac{1}{a}} \right]$

$ \Rightarrow $$a = 1$.

$x + y + z = 15$ if $9,\;x,\;y,\;z,\;a$ are in $A.P.$; while $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3}$ if $9,\;x,\;y,\;z,\;a$ are in $H.P.$, then the value of $a$ will be

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