Question

\[x + y + z = 15\] 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
A. 1
B. 2
C. 3
D. 9

Answer 1

Hint:
For the aforementioned question, we will apply the idea that if two terms of an A.P are "a" and "b," and there are "n" arithmetic means between them, such as \[{x_1},{x_2},{x_3}........{x_n}\], then their sum is equal to n times the average value of "a" and "b."
\[{x_1},{x_2},{x_3}........{x_n} = n\left( {\frac{{a + b}}{2}} \right)\]. We also know that, if \[{a_1},{a_2},{a_3}........\]is in Harmonic progression, and then \[\left( {\frac{1}{{{a_1}}},\frac{1}{{{a_2}}},\frac{1}{{{a_3}}}} \right).........\]must be in Arithmetic progression.
Formula used:
\[{x_1},{x_2},{x_3}........{x_n}\], then their sum is equal to n times the average value of "a" and "b."
\[{x_1},{x_2},{x_3}........{x_n} = n\left( {\frac{{a + b}}{2}} \right)\]. We also know that, if \[{a_1},{a_2},{a_3}........\]is in Harmonic progression, and then \[\left( {\frac{1}{{{a_1}}},\frac{1}{{{a_2}}},\frac{1}{{{a_3}}}} \right).........\]must be in Arithmetic progression.

Complete step-by-step solution:
We have been given that if \[9,x,y,z,a\] 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
If there are n arithmetic means, such as \[{x_1},{x_2},{x_3}........{x_n}\]between ‘a’ and ‘b’,
We know that,
\[{x_1},{x_2},{x_3}........{x_n} = n\left( {\frac{{a + b}}{2}} \right)\]
Between ‘a’ and ‘b’, there are three arithmetic means:
\[x,{\rm{ }}y\] and \[z\]
Let us form the equation:
\[x + y + z = 15 \ldots {\rm{ (i)}}\]
\[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{5}{3} \ldots \ldots ..{\rm{ (ii) }}\]
If\[{\rm{a}},{\rm{x}},{\rm{y}},{\rm{z}}\]are in A.P
Then, \[x,y,z\] are in A.P
Therefore, the equation becomes,
\[2{\rm{y}} = {\rm{x}} + {\rm{z}} \ldots \ldots .(iii)\]
\[ \Rightarrow 2{\rm{y}} + {\rm{y}} = 15 \ldots \ldots \ldots .\]Comparing equations (iii) and (i))
Thus, we have
\[ \Rightarrow 3{\rm{y}} = 15 \Rightarrow {\rm{y}} = 5\]
Now, we have to substitute three arithmetic means in the above formula:
\[ \Rightarrow x + y + z = 3\left( {\frac{{a + b}}{2}} \right)\]
Substitute the values of the arithmetic mean
\[ \Rightarrow 15 = 3\left( {\frac{{a + b}}{2}} \right)\]
Now we have to group all similar terms by having constants one side and variables on other side: we get
\[ \Rightarrow \frac{{15 \times 2}}{3} = a + b\]
On simplifying the above equation, we get
\[ \Rightarrow 10 = a + b\]
Rewrite the above expression as,
\[ \Rightarrow a + b = 10 \ldots \ldots \ldots (vi)\]
Now, if \[a,x,y,z\;\]are in HP, then \[x,y,z\] in H.P
\[\therefore \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \ldots \ldots {\rm{ (iv) }}\]
\[ \Rightarrow \frac{1}{{\rm{x}}} + \frac{1}{{\rm{y}}} + \frac{1}{{\rm{z}}} = \frac{5}{3} \ldots \ldots .({\rm{v}})\]
Now, from the equations (iv) and (v) we can write
\[\frac{2}{{\rm{y}}} + \frac{1}{{\rm{y}}} = \frac{5}{3} = \frac{5}{3}\]
Putting the equation (iv) in equation (v)
\[\frac{3}{y} = \frac{5}{3}\]
Now, simplify the above by solving it for\[y\]:
\[ \Rightarrow y = \frac{9}{5}\]
Also, a, y, b are in H.P.
\[\frac{1}{{\rm{a}}} + \frac{1}{{\;{\rm{b}}}} = \frac{2}{{\rm{y}}}\]
\[ \Rightarrow \frac{{{\rm{a}} + {\rm{b}}}}{{{\rm{ab}}}} = \frac{{ - 2}}{{\rm{y}}} = \frac{{2 \times 5}}{9}\]
\[ \Rightarrow \frac{{{\rm{a}} + {\rm{b}}}}{{{\rm{ab}}}} = \frac{{10}}{9} \ldots \ldots .({\rm{vi}})\]
By comparing (vi) and (vii) we can say that
\[\frac{{{\rm{a}} + {\rm{b}}}}{{{\rm{ab}}}} = \frac{{10}}{9}\]
Now substitute the value of\[a + b\], we get
\[ \Rightarrow \frac{{10}}{{ab}} = \frac{{10}}{9}\]
\[ \Rightarrow ab = 9\]………. (viii)
Now, we have to substitute the value of\[b\], we get
\[ \Rightarrow a(10 - a) = 9\]
Multiply the terms with terms inside the parentheses, we get
\[ \Rightarrow 10a - {a^2} = 9\]
Rewrite the equation with quadratic formula:
\[ \Rightarrow {a^2} - 10a + 9 = 0\]
Now, we have to factor the above equation, we get
\[(a - 1)(a - 9) = 0\]
Now, we have to equate the factors to\[0\]:
\[a = 1\] Or \[a = 9\]
Therefore, the value becomes; we get
\[a \times b = 9 \times 1 = 1 \times 9 = 9\]
Therefore, the value of \[a\] will be \[9\].
Hence, the option D is correct.
Note:
Students should keep in mind that the term of H.P has an opposite in A.P. Remember that Arithmetic Progression is a constant sequence of numbers, and its short form is (A.P), whereas Harmonic Progression is abbreviated as H.P.