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8. Sequences and Series
easy
अनंत गुणोत्तर श्रेणी $\frac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}},\frac{1}{{2 - \sqrt 2 }},\frac{1}{2}.....$ के पदों का योग होगा
A
$\sqrt 2 {(\sqrt 2 + 1)^2}$
B
${(\sqrt 2 + 1)^2}$
C
$5\sqrt 2 $
D
$3\sqrt 2 + \sqrt 5 $
Solution
(a) $\frac{{\sqrt 2 + 1}}{{\sqrt 2 – 1}},\frac{1}{{\sqrt 2 (\sqrt 2 – 1)}},\frac{1}{2},……$
श्रेणी का सार्वअनुपात $ = \frac{1}{{\sqrt 2 (\sqrt 2 + 1)}}$
$\therefore $ sum $ = \frac{a}{{1 – r}} = \frac{{\left( {\frac{{\sqrt 2 + 1}}{{\sqrt 2 – 1}}} \right)}}{{\left( {1 – \frac{1}{{\sqrt 2 (\sqrt 2 + 1)}}} \right)}}$
$ = \frac{{(\sqrt 2 + 1)}}{{(\sqrt 2 – 1)}}.\,\frac{{\sqrt 2 \,(\sqrt 2 + 1)}}{{(1 + \sqrt 2 )}}$$ = \sqrt 2 {(\sqrt 2 + 1)^2}$.
Standard 11
Mathematics
