
If \[{a^2},{b^2},{c^2}\] are in A.P., then \[{(b + c)^{ - 1}},{(c + a)^{ - 1}}\] and \[{(a + b)^{ - 1}}\] will be in
A. H.P.
B. G.P.
C. A.P.
D. None of these
Answer
164.4k+ views
Hint:
In this question, we are given an AP of three numbers and must use it to demonstrate that the other three numbers are also in AP. For this, we will assume that the difference between the second and first terms of a given AP is equal to the difference between the third and second terms of the same AP. We will use this to demonstrate our new AP. We will use the property \[{a^2} - {b^2} = (a + b)(a - b)\].
Formula used:
\[{a},{b},{c}\] are in Arithmetic progression.
\[2b = a + c\]
Complete step-by-step solution:
We have been given that \[{a^2},{b^2},{c^2}\] are in Arithmetic progression.
Therefore, it can be written as
\[2{b^2} = {a^2} + {c^2}\]
The above equation can also be written as,
\[ \Rightarrow {b^2} - {a^2} = {c^2} - {b^2}\]
Expand the above equation using square rule, we get
\[ \Rightarrow (b + a)(b - a) = (c - b)(c + b)\]
Now, we have to restructure the above equation as fraction, we obtain
\[ \Rightarrow \frac{{b - a}}{{c + b}} = \frac{{c - b}}{{b + a}}\]
Now, we have to multiply either side of the equation’s denominator by \[(c + a)\]\[ \Rightarrow \frac{{b - a}}{{(c + a)(c + b)}} = \frac{{c - b}}{{(b + a)(c + a)}}\]
Multiplying both sides of the above equation by \[\frac{1}{{c + a}}\]:
\[ \Rightarrow \frac{1}{{c + a}} - \frac{1}{{b + c}} = \frac{1}{{a + b}} - \frac{1}{{c + a}}\]
Now, we know that \[\frac{1}{{b + c}},\frac{1}{{c + a}},\frac{1}{{a + b}}\] are in A.P.
The above obtained equation can also be written as inverse form:
\[{(b + c)^{ - 1}},{(c + a)^{ - 1}},{(a + b)^{ - 1}}\]
Therefore, if \[{a^2},{b^2},{c^2}\] are in A.P., then \[{(b + c)^{ - 1}},{(c + a)^{ - 1}}\]and\[{(a + b)^{ - 1}}\] will be in A.P
Hence, the option C is correct.
Note:
Because the calculations and equation is complex, students must exercise caution at all times. In these questions, students may make mistakes with the signs. When using \[{a^2} - {b^2} = (a + b)(a - b)\], ensure that the negative sign is with b on both sides. It is sufficient to determine that the difference between the third and second terms is equal to the difference between the second and first terms.
In this question, we are given an AP of three numbers and must use it to demonstrate that the other three numbers are also in AP. For this, we will assume that the difference between the second and first terms of a given AP is equal to the difference between the third and second terms of the same AP. We will use this to demonstrate our new AP. We will use the property \[{a^2} - {b^2} = (a + b)(a - b)\].
Formula used:
\[{a},{b},{c}\] are in Arithmetic progression.
\[2b = a + c\]
Complete step-by-step solution:
We have been given that \[{a^2},{b^2},{c^2}\] are in Arithmetic progression.
Therefore, it can be written as
\[2{b^2} = {a^2} + {c^2}\]
The above equation can also be written as,
\[ \Rightarrow {b^2} - {a^2} = {c^2} - {b^2}\]
Expand the above equation using square rule, we get
\[ \Rightarrow (b + a)(b - a) = (c - b)(c + b)\]
Now, we have to restructure the above equation as fraction, we obtain
\[ \Rightarrow \frac{{b - a}}{{c + b}} = \frac{{c - b}}{{b + a}}\]
Now, we have to multiply either side of the equation’s denominator by \[(c + a)\]\[ \Rightarrow \frac{{b - a}}{{(c + a)(c + b)}} = \frac{{c - b}}{{(b + a)(c + a)}}\]
Multiplying both sides of the above equation by \[\frac{1}{{c + a}}\]:
\[ \Rightarrow \frac{1}{{c + a}} - \frac{1}{{b + c}} = \frac{1}{{a + b}} - \frac{1}{{c + a}}\]
Now, we know that \[\frac{1}{{b + c}},\frac{1}{{c + a}},\frac{1}{{a + b}}\] are in A.P.
The above obtained equation can also be written as inverse form:
\[{(b + c)^{ - 1}},{(c + a)^{ - 1}},{(a + b)^{ - 1}}\]
Therefore, if \[{a^2},{b^2},{c^2}\] are in A.P., then \[{(b + c)^{ - 1}},{(c + a)^{ - 1}}\]and\[{(a + b)^{ - 1}}\] will be in A.P
Hence, the option C is correct.
Note:
Because the calculations and equation is complex, students must exercise caution at all times. In these questions, students may make mistakes with the signs. When using \[{a^2} - {b^2} = (a + b)(a - b)\], ensure that the negative sign is with b on both sides. It is sufficient to determine that the difference between the third and second terms is equal to the difference between the second and first terms.
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