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a,b,c in HP: a/(b+c), b/(c+a), c/(a+b) Progression | JEE

JEE Maths question with a full step-by-step solution.

Question
If a,b,ca, b, c are in H.P., then ab+c, bc+a, ca+b\frac{a}{b+c},\ \frac{b}{c+a},\ \frac{c}{a+b} are in
AA.P.
BH.P.correct
CG.P.
DNone of these
Solution
Step 1: a,b,ca,b,c in H.P. ⇒1a,1b,1c\Rightarrow\dfrac1a,\dfrac1b,\dfrac1c in A.P. Step 2: Multiply this A.P. by (a+b+c)(a+b+c) and subtract 11 from each term:
a+b+ca−1, a+b+cb−1, a+b+cc−1=b+ca, c+ab, a+bc are in A.P.\frac{a+b+c}{a}-1,\ \frac{a+b+c}{b}-1,\ \frac{a+b+c}{c}-1=\frac{b+c}{a},\ \frac{c+a}{b},\ \frac{a+b}{c}\ \text{are in A.P.}
 Step 3: Their reciprocals are therefore in H.P., so ab+c,bc+a,ca+b\dfrac{a}{b+c},\dfrac{b}{c+a},\dfrac{c}{a+b} are in H.P. Correct answer: (2)
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