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8. Sequences and Series
hard
If ${a^2},\;{b^2},\;{c^2}$ are in $A.P.$, then ${(b + c)^{ - 1}},\;{(c + a)^{ - 1}}$ and ${(a + b)^{ - 1}}$ will be in
A
$H.P.$
B
$G.P.$
C
$A.P.$
D
None of these
Solution
(c) ${a^2},\;{b^2},\;{c^2}$ are in $A.P.$
Therefore ${a^2} + (ab + bc + ca)$, ${b^2} + (ab + bc + ca)$, ${c^2} + (ab + bc + ca)$ will be in $A.P.$
$ \Rightarrow $ $\{ a(a + b) + c\,(a + b)\} ,\;\{ b(b + a) + c\,(b + a)\} ,\;$
$c(c + b) + a(b + c)$ will be in $A.P.$
==>$(a + b)(a + c),\;(b + a)(b + c),\;(c + a)(c + b)$ will be in $A.P.$
$ \Rightarrow $$\frac{1}{{b + c}},\;\frac{1}{{c + a}},\;\frac{1}{{a + b}}$ will be in $A.P.$
{ Dividing each term by $(a+b)(b+c)(c+a)$ }
Standard 11
Mathematics
