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If ${a_1},\;{a_2},............,{a_n}$ are in $A.P.$ with common difference , $d$, then the sum of the following series is $\sin d(\cos {\rm{ec}}\,{a_1}.co{\rm{sec}}\,{a_2} + {\rm{cosec}}\,{a_2}.{\rm{cosec}}\,{a_3} + ...........$$ + {\rm{cosec}}\;{a_{n - 1}}{\rm{cosec}}\;{a_n})$
$\sec {a_1} - \sec {a_n}$
$\cot {a_1} - \cot {a_n}$
$\tan {a_1} - \tan {a_n}$
$c{\rm{osec}}\;{a_1} - {\rm{cosec}}\;{a_n}$
Solution
(b) As given $d = {a_2} – {a_1} = {a_3} – {a_2} = …. = {a_n} – {a_{n – 1}}$
$\therefore $ $\sin d\,\{ co{\rm{sec}}\;{a_1}co{\rm{sec}}\;{a_2} + ….. + {\rm{cosec}}\;{a_{n – 1}}{\rm{cosec}}\;{a_n}\} $
$ = \frac{{\sin ({a_2} – {a_1})}}{{\sin {a_1}.\;\sin {a_2}}} + …… + \frac{{\sin ({a_n} – {a_{n – 1}})}}{{\sin {a_{n – 1}}\sin {a_n}}}$
$ = (\cot {a_1} – \cot {a_2}) + (\cot {a_2} – \cot {a_3}) + …. + (\cot {a_{n – 1}} – \cot {a_n})$
$ = \cot {a_1} – \cot {a_n}$.
