Question

If \[{a^2},{b^2},{c^2}\] be in A.P., then \[\frac{a}{{b + c}},\frac{b}{{c + a}},\frac{c}{{a + b}}\] will be in
A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer 1

Hint:
We are given an AP of three numbers in this question. We will assume for this that the difference between the third term and the second term of the given AP is equal to the difference between the second term and the first term of the given AP. We'll apply the formula \[{a^2} - {b^2} = (a + b)(a - b)\].
Formula use:
The arithmetic progression is \[a,b,c\]
\[(b - a) = (c - b)\]
Complete step-by-step solution
Three numbers in arithmetic progression \[{a^2},{b^2},{c^2}\] are given to us in this instance. That \[\frac{a}{{b + c}},\frac{b}{{c + a}},\frac{c}{{a + b}}\]are also in the arithmetic progression must be demonstrated.
The difference between the second term and the first term will equal the difference between the third term and the second term because the numbers \[{a^2},{b^2},{c^2}\] are in arithmetic progression.
Thus, we obtain
 \[{b^2} - {a^2} = {c^2} - {b^2}\]
We already know that,
\[{a^2} - {b^2} = (a + b)(a - b)\]
So the above equation become,
\[(b - a)(b + a) = (c - b)(c + b)\]
On dividing both sides of the equation by \[(b + a)(c + b)\], we obtain
\[ \Rightarrow \frac{{(b - a)(b + a)}}{{(b + a)(c + b)}} = \frac{{(c - b)(c + b)}}{{(b + a)(c + b)}}\]
Now, we have to cancel \[(b + a)\] on left side and \[({\rm{c}} + {\rm{b}})\] on right side, we
\[ \Rightarrow \frac{{(b - a)}}{{(c + b)}} = \frac{{(c - b)}}{{(b + a)}}\]
On multiplying both sides by\[\frac{1}{{(c + a)}}\]we get:
\[ \Rightarrow \frac{{(b - a)}}{{(c + b)(c + a)}} = \frac{{(c - b)}}{{(b + a)(c + a)}}\]
For proper symmetry, let's add and subtract "a" from the right side's numerator and "c" from the left side's numerator.
This gives us:
\[ \Rightarrow \frac{{(b + c - c - a)}}{{(c + b)(c + a)}} = \frac{{(c + a - a - b)}}{{(b + a)(c + a)}}\]
Using common negative signs, we obtain:
\[ \Rightarrow \frac{{(b + c) - (c + a)}}{{(c + b)(c + a)}} = \frac{{(c + a) - (a + b)}}{{(b + a)(c + a)}}\]
Now that we have terms on both sides separated, we have:
\[ \Rightarrow \frac{{(b + c)}}{{(c + b)(c + a)}} - \frac{{(c + a)}}{{(c + b)(c + a)}} = \frac{{(c + a)}}{{(b + a)(c + a)}} - \frac{{(a + b)}}{{(b + a)(c + a)}}\]
Now, cancelling the common terms, we get:
\[ \Rightarrow \frac{1}{{(c + a)}} - \frac{1}{{(c + b)}} = \frac{1}{{(a + b)}} - \frac{1}{{(c + a)}}\]
Multiply both sides by \[(a + b + c)\]:
\[ \Rightarrow (a + b + c)\left( {\frac{1}{{c + a}} - \frac{1}{{c + b}}} \right) = (a + b + c)\left( {\frac{1}{{a + b}} - \frac{1}{{c + a}}} \right)\]
We get,
\[ \Rightarrow \frac{{a + b + c}}{{(c + a)}} - \frac{{a + b + c}}{{(c + b)}} = \frac{{a + b + c}}{{(a + b)}} - \frac{{a + b + c}}{{(c + a)}}\]
\[ \Rightarrow \frac{{(c + a) + b}}{{(c + a)}} - \frac{{a + (c + b)}}{{(c + b)}} = \frac{{(a + b) + c}}{{(a + b)}} - \frac{{(c + a) + b}}{{(c + a)}}\]
Now, separate the terms, we get:
\[ \Rightarrow \frac{{(c + a)}}{{(c + a)}} + \frac{b}{{(c + a)}} - \frac{a}{{(c + b)}} - \frac{{(c + b)}}{{(c + b)}} = \frac{{(a + b)}}{{(a + b)}} + \frac{c}{{(a + b)}} - \frac{{(c + a)}}{{(c + a)}} - \frac{b}{{(c + a)}}\]
Now, we have to cancel out common terms, we get:
\[ \Rightarrow 1 + \frac{b}{{(c + a)}} - \frac{a}{{(c + b)}} - 1 = 1 + \frac{c}{{(a + b)}} - 1 - \frac{b}{{(c + a)}}\]
Now simplify, we get
\[ \Rightarrow \frac{b}{{(c + a)}} - \frac{a}{{(c + b)}} = \frac{c}{{(a + b)}} - \frac{b}{{(c + a)}}\]
According to the equation above, the difference between the second and first terms should be equal to the difference between the third and second terms if \[\frac{a}{{(c + b)}}\] were to be the first term, \[\frac{b}{{(c + a)}}\]the second term, and \[\frac{c}{{(a + b)}}\]the third term. As a result \[\frac{a}{{(b + c)}},\frac{b}{{(c + a)}},\frac{c}{{(a + b)}}\] are therefore in arithmetic progression.
Hence, the option A is correct.
Note:
Due to the complexity of the calculations and equation, students should exercise caution when carrying out each step. Signs in these questions are subject to student error. As you apply the formula \[{a^2} - {b^2} = (a + b)(a - b)\], be sure that the negative sign is on both sides of the \[b\].