
\[x + y + z = 15\] 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
162.6k+ views
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.
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.
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