
If \[{b^2},{a^2},{c^2}\] are in A.P., then \[a + b,b + c,c + a\] will be in
A. A.P.
B. G.P.
C. H.P.
D. None of these
Answer
163.2k+ 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 A.P.
\[2b = a+c\].
Complete step-by-step solution:
We are given three numbers in arithmetic progression \[{b^2},{a^2},{c^2}\].
An arithmetic progression is a number sequence in which the difference between any two successive members is a constant, whereas a harmonic progression is a term sequence in which the reciprocal of the terms is in A.P.
We have to assume that \[a + b,b + c,c + a\] are in A.P.
So we must demonstrate that \[{b^2},{a^2},{c^2}\] must be in A.P.
Because \[{b^2},{a^2},{c^2}\] are in arithmetic progression, the difference between the second and first terms is equal to the difference between the third and second terms.
As a result, \[{a^2} - {b^2} = {c^2} - {a^2}\]is obtained.
Or, we can write it as,
\[(a - b)(a + b) = (c - a)(c + a)\]
The above equation can also be written in fraction form as,
\[\frac{{a - b}}{{c + a}} = \frac{{c - a}}{{a + b}}\]
Expand the above equation and write as,
\[\frac{{(a + c) - (b + c)}}{{(c + a)(b + c)}} = \frac{{(b + c) - (a + b)}}{{(a + b)(b + c)}}\]
Now, cancel the similar terms to make it simpler, we get
\[\frac{1}{{b + c}} - \frac{1}{{c + a}} = \frac{1}{{a + b}} - \frac{1}{{b + c}}\]
Let’s restructure the above equation explicitly by having similar terms on one side:\[\frac{1}{{b + c}} - \frac{1}{{a + b}} = \frac{1}{{c + a}} - \frac{1}{{b + c}}\]
That implies,\[a + b,b + c,c + a\] are in Harmonic Progression.
Therefore, if \[{b^2},{a^2},{c^2}\] are in A.P., then \[a + b,b + c,c + a\] will be in H.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.
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 A.P.
\[2b = a+c\].
Complete step-by-step solution:
We are given three numbers in arithmetic progression \[{b^2},{a^2},{c^2}\].
An arithmetic progression is a number sequence in which the difference between any two successive members is a constant, whereas a harmonic progression is a term sequence in which the reciprocal of the terms is in A.P.
We have to assume that \[a + b,b + c,c + a\] are in A.P.
So we must demonstrate that \[{b^2},{a^2},{c^2}\] must be in A.P.
Because \[{b^2},{a^2},{c^2}\] are in arithmetic progression, the difference between the second and first terms is equal to the difference between the third and second terms.
As a result, \[{a^2} - {b^2} = {c^2} - {a^2}\]is obtained.
Or, we can write it as,
\[(a - b)(a + b) = (c - a)(c + a)\]
The above equation can also be written in fraction form as,
\[\frac{{a - b}}{{c + a}} = \frac{{c - a}}{{a + b}}\]
Expand the above equation and write as,
\[\frac{{(a + c) - (b + c)}}{{(c + a)(b + c)}} = \frac{{(b + c) - (a + b)}}{{(a + b)(b + c)}}\]
Now, cancel the similar terms to make it simpler, we get
\[\frac{1}{{b + c}} - \frac{1}{{c + a}} = \frac{1}{{a + b}} - \frac{1}{{b + c}}\]
Let’s restructure the above equation explicitly by having similar terms on one side:\[\frac{1}{{b + c}} - \frac{1}{{a + b}} = \frac{1}{{c + a}} - \frac{1}{{b + c}}\]
That implies,\[a + b,b + c,c + a\] are in Harmonic Progression.
Therefore, if \[{b^2},{a^2},{c^2}\] are in A.P., then \[a + b,b + c,c + a\] will be in H.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.
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