
If \[x,1,z\] are in A.P. and \[x,2,z\] are in G.P., then by which progress \[x,4,z\] are in?
A. A.P.
B. G.P.
c. H.P.
D. None of these
Answer
162.9k+ views
Hint: Use the conditions for three terms being in A.P. and G.P. Then find the relationship of the terms \[x,4,z\] using these two conditions.
Formula used
If three terms \[a,b,c\] are in A.P., then \[2b = a + c\]
If three terms \[a,b,c\] are in G.P., then \[{b^2} = ac\]
If three terms \[a,b,c\] are in H.P., then \[b = \dfrac{{2ac}}{{a + c}}\]
Complete step by step solution
Given that \[x,1,z\] are in A.P.
So, \[2 = x + z - - - - - \left( i \right)\]
Also given that \[x,2,z\] are in G.P.
So, \[4 = xz - - - - - \left( {ii} \right)\]
Dividing equation \[\left( i \right)\] by equation \[\left( {ii} \right)\], we get
\[\dfrac{2}{4} = \dfrac{{x + z}}{{xz}}\]
\[ \Rightarrow \dfrac{1}{2} = \dfrac{{x + z}}{{xz}}\]
Take reciprocals.
\[ \Rightarrow 2 = \dfrac{{xz}}{{x + z}}\]
Multiply both sides by \[2\]
\[ \Rightarrow 4 = \dfrac{{2xz}}{{x + z}}\]
So, it is seen that \[x,4,z\] are in H.P.
Hence option C is correct.
Note: Many students can’t remember all the three conditions properly. Always remember that three terms will be in A.P. if the twice of the middle term is equal to the sum of the other two. Three terms will be in G.P. if the square of the middle term is equal to the product of the other two. Three terms will be in H.P. if the reciprocals of the terms are in A.P.
Formula used
If three terms \[a,b,c\] are in A.P., then \[2b = a + c\]
If three terms \[a,b,c\] are in G.P., then \[{b^2} = ac\]
If three terms \[a,b,c\] are in H.P., then \[b = \dfrac{{2ac}}{{a + c}}\]
Complete step by step solution
Given that \[x,1,z\] are in A.P.
So, \[2 = x + z - - - - - \left( i \right)\]
Also given that \[x,2,z\] are in G.P.
So, \[4 = xz - - - - - \left( {ii} \right)\]
Dividing equation \[\left( i \right)\] by equation \[\left( {ii} \right)\], we get
\[\dfrac{2}{4} = \dfrac{{x + z}}{{xz}}\]
\[ \Rightarrow \dfrac{1}{2} = \dfrac{{x + z}}{{xz}}\]
Take reciprocals.
\[ \Rightarrow 2 = \dfrac{{xz}}{{x + z}}\]
Multiply both sides by \[2\]
\[ \Rightarrow 4 = \dfrac{{2xz}}{{x + z}}\]
So, it is seen that \[x,4,z\] are in H.P.
Hence option C is correct.
Note: Many students can’t remember all the three conditions properly. Always remember that three terms will be in A.P. if the twice of the middle term is equal to the sum of the other two. Three terms will be in G.P. if the square of the middle term is equal to the product of the other two. Three terms will be in H.P. if the reciprocals of the terms are in A.P.
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