English
Hindi
8. Sequences and Series
hard

$a_1, a_2, a_3, ….a_n$ સમાંતર શ્રેણીમાં છે. જો તેનો સામાન્ય તફાવત $d$ હોય, તો $sin\,\, d[cosec\ a_1 . cosec\ a_2 + cosec\ a_2 . cosec\ a_3 +….+cosec\ a_{n -1} . cosec\ a_n] $ ની કિમત મેળવો.

A

$cosec\ a_1 - cosec\ a_n$

B

$sec\ a_1 - sec\ a_n$

C

$cot\ a_1 - cot\ a_n$

D

$tan\ a_1 - tan\ a_n$

Solution

$a_1, a_2, a_3, ….a_n$ સમાંતર શ્રેણીમાં છે.

$d = a_2 – a_1 = a_3 – a_2 = …..= a_n – a_{n-1}$

હવે,$sin\,\, d[cosec\ a_1 . cosec\ a_2 + cosec\ a_2 . cosec\ a_3 +….+cosec\ a_{n -1} . cosec\ a_n]$

$=\frac{\sin ({{a}_{2}}-{{a}_{1}})}{\sin {{a}_{1}}\sin {{a}_{2}}}+\frac{\sin ({{a}_{3}}-{{a}_{2}})}{\sin {{a}_{2}}\sin {{a}_{3}}}+…+\frac{\sin ({{a}_{n}}-{{a}_{n-1}})}{\sin {{a}_{n-1}}\sin {{a}_{n}}}\,\,\,$

$=\frac{\sin {{a}_{2}}\cos {{a}_{1}}-\cos {{a}_{2}}\sin {{a}_{1}}}{\sin {{a}_{1}}\sin {{a}_{2}}}+….+\frac{\sin {{a}_{n}}{{\cos }_{n-1}}\cos {{a}_{n}}\sin {{a}_{n-1}}}{\sin {{a}_{n-1}}\sin {{a}_{n}}}$

$=\cot {{a}_{1}}-\cot {{a}_{2}}+\cot {{a}_{2}}-\cot {{a}_{3}}+…+\cot {{a}_{n-1}}-\cot {{a}_{n}}$

$=\cot {{a}_{1}}-\cot {{a}_{n}}$

Standard 11
Mathematics

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