ifa,b,c are in h.p , ab+bc+ca=15 then ca =?
Answers
Answered by
17
Answer:
Lets divide the whole equation by abcabc we get \dfrac{1}{a}+\dfrac{1}{b}+\dfrac1c=\dfrac{15}{abc}
a
1
+
b
1
+
c
1
=
abc
15
Now we know \dfrac1a +\dfrac1c = \dfrac2b \quad \quad [HP]
a
1
+
c
1
=
b
2
[HP]
\dfrac3b=\dfrac{15}{abc}
b
3
=
abc
15
ac=5ac=5
#Capricorn Answers
Answered by
1
Answer:
Step-by-step explanation:
Since a,b,c are in harmonic progression, their
reciprocals form an arithmetic sequence:
Let d be the common difference. Then
Solve each for d
Since both equal d, they are equal to each other:
Multiply through by LCD of abc
ac - bc = ab - ac
2ac = ab + bc
And since we are given that
ab + bc + ca = 15
We can substitute 2ac for ab + bc and get
2ac + ca = 15
And since ac is the same as ca
2ca + ca = 15
3ca = 15
ca = 5
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