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8. Sequences and Series
medium
If $x,y,z$ are in $A.P. $ and ${\tan ^{ - 1}}x,{\tan ^{ - 1}}y$ and ${\tan ^{ - 1}}z$ are also in $A.P.$, then
A
$x = y = z$
B
$x = y = - z$
C
$x = 1;y = 2;z = 3$
D
$x = 2;y = 4;z = 6$
Solution
(a) $2{\tan ^{ – 1}}y = {\tan ^{ – 1}}x + {\tan ^{ – 1}}z$
==> ${\tan ^{ – 1}}\left( {\frac{{2y}}{{1 – {y^2}}}} \right) = {\tan ^{ – 1}}\left( {\frac{{x + z}}{{1 – xz}}} \right)$
==> $\frac{{2y}}{{1 – {y^2}}} = \frac{{x + z}}{{1 – xz}}$
But $2y = x + z$
$1 – {y^2} = 1 – xz$
==> ${y^2} = xz$
$xyz$ are both in $G.P. $ and $A.P.$,
$x = y = z$.
Standard 11
Mathematics
