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Question: If \[\dfrac{a}{b+c}\], \[\dfrac{b}{a+c}\], \[\dfrac{c}{a+b}\] are in AP, then A. \[a,b,c\] are in ...

If ab+c\dfrac{a}{b+c}, ba+c\dfrac{b}{a+c}, ca+b\dfrac{c}{a+b} are in AP, then
A. a,b,ca,b,c are in AP
B. c,a,bc,a,b are in AP
C. a2,b2,c2{{a}^{2}},{{b}^{2}},{{c}^{2}}are in AP
D. a,b,ca,b,c are in GP.

Explanation

Solution

In this question the first step is to apply the condition of AP series because in the question it is given that the series is in AP. Now simplify the obtained expression from the condition. After simplifying, observe the pattern followed by the expression. Now you can solve the problem.

Complete step by step answer:
If the given sequence is an AP series or AP sequence then it must follow the condition for the AP series and that condition of the series is that the difference between the consecutive terms of the series must be equal. So, applying the condition of the AP series in the given sequence then we get,
The given series is ab+c\dfrac{a}{b+c}, ba+c\dfrac{b}{a+c}, ca+b\dfrac{c}{a+b}
We know the condition of the AP series i.e. a2−a1=a3−a2{{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}
Where,
a1={{a}_{1}}= first term of the series
a2={{a}_{2}}= second term of the series
a3={{a}_{3}}= third term of the series
Now substituting all the values in the condition, we get
⇒ba+c−ab+c=ca+b−ba+c\Rightarrow \dfrac{b}{a+c}-\dfrac{a}{b+c}=\dfrac{c}{a+b}-\dfrac{b}{a+c}
By taking the LCM on both the sides, we get
⇒b(b+c)−a(a+c)(a+c)(b+c)=c(a+c)−b(a+b)(a+b)(a+c)\Rightarrow \dfrac{b(b+c)-a(a+c)}{(a+c)(b+c)}=\dfrac{c(a+c)-b(a+b)}{(a+b)(a+c)}
Cancel out (a+c)(a+c) from both the sides of denominator, we get
⇒b2+bc−a2−acb+c=ac+c2−ab−b2a+b\Rightarrow \dfrac{{{b}^{2}}+bc-{{a}^{2}}-ac}{b+c}=\dfrac{ac+{{c}^{2}}-ab-{{b}^{2}}}{a+b}
Rearrange the numerator of both the sides, we get
⇒b2−a2+bc−acb+c=ac−ab+c2−b2a+b\Rightarrow \dfrac{{{b}^{2}}-{{a}^{2}}+bc-ac}{b+c}=\dfrac{ac-ab+{{c}^{2}}-{{b}^{2}}}{a+b}
Now simplify the above expression by taking the common multiples and by using the identity
⇒a2−b2=(a+b)(a−b)\Rightarrow {{a}^{2}}-{{b}^{2}}=(a+b)(a-b)
⇒(b+a)(b−a)+c(b−a)b+c=a(c−b)+(c+b)(c−b)a+b\Rightarrow \dfrac{(b+a)(b-a)+c(b-a)}{b+c}=\dfrac{a(c-b)+(c+b)(c-b)}{a+b}
Again taking the common multiples, we get
⇒(b−a)(b+a+c)b+c=(c−b)(a+b+c)a+b\Rightarrow \dfrac{(b-a)(b+a+c)}{b+c}=\dfrac{(c-b)(a+b+c)}{a+b}
Again cancelling out the common term of the numerator i.e. (a+b+c)(a+b+c), then we get
⇒b−ab+c=c−ba+b\Rightarrow \dfrac{b-a}{b+c}=\dfrac{c-b}{a+b}
Now cross multiply the above expression,
⇒(b−a)(b+a)=(c−b)(c+b)\Rightarrow (b-a)(b+a)=(c-b)(c+b)
By simplifying the expression, we get
⇒b2−a2=c2−b2\Rightarrow {{b}^{2}}-{{a}^{2}}={{c}^{2}}-{{b}^{2}}
From the above expression we can conclude that if the series of numbers a2,b2,c2{{a}^{2}},{{b}^{2}},{{c}^{2}} is given then the condition for an AP series i.e. the difference between the two consecutive numbers is equal. And the condition for the series a2,b2,c2{{a}^{2}},{{b}^{2}},{{c}^{2}}is b2−a2=c2−b2{{b}^{2}}-{{a}^{2}}={{c}^{2}}-{{b}^{2}}. So we can say that a2,b2,c2{{a}^{2}},{{b}^{2}},{{c}^{2}}are in AP. And the same result we are getting by solving the given question.
Hence we can conclude that if ab+c\dfrac{a}{b+c}, ba+c\dfrac{b}{a+c}, ca+b\dfrac{c}{a+b} are in AP, then a2,b2,c2{{a}^{2}},{{b}^{2}},{{c}^{2}}numbers also forms an AP series.

So, the correct answer is “Option C”.

Note: Let’s have a look at the GP series. Geometric Progression or GP is a sequence or a series of numbers in which the ratio between the two consecutive terms is the same. And if we take the reciprocal of the arithmetic progression series then the series obtained is known as harmonic series or HP.