English
Hindi
8. Sequences and Series
normal

જો એક સમગુણોતર શ્રેણીના પ્રથમ $n$ પદનો સરવાળો $S$,ગુણાકાર $P$ અને શ્રેણીના પ્રથમ $n$ પદનાં વ્યસ્તનો સરવાળો $R$ હોય તો $P^2 = ……$

A

${\left( {\frac{S}{R}} \right)^n}$

B

$\left( {\frac{S}{R}} \right)$

C

${\left( {\frac{R}{S}} \right)^n}$

D

$\frac{R}{S}$

Solution

અહીં $S = a + ar + ar^2 + … + ar^{n-1}  …. (1)$

અને $P = a \times (ar) \times (ar^2) \times … \times (ar^{n-1})$

$\therefore \,\,P\,\,=\,\,{{a}^{n}}\,{{r}^{1\,+\,2\,+\,..\,+\,n\,-1}}\,\,\,=\,\,{{a}^{n}}\,{{r}^{\Sigma (n-1)}}$

$=\,\,{{a}^{n}}{{r}^{\frac{n}{2}\,(n+1)-n}}\,=\,\,\,\,{{a}^{n}}\,{{r}^{\frac{n}{2}\,(n\,-\,1)}}\,…….\,(2)$

$\therefore \,\,{{P}^{2}}\,\,=\,\,{{a}^{2n}}{{r}^{n(n-1)}}$

વળી $R\,\,=\,\,\frac{1}{a}\,+\,\frac{1}{ar}\,+\,\frac{1}{a{{r}^{2}}}\,+\,…\,+\,\frac{1}{a{{r}^{n-1}}}$

$=\,\,{{a}^{-1}}\,+\,{{a}^{-1}}\,{{r}^{-1}}\,+\,{{a}^{-1}}\,{{r}^{-2}}\,+\,…\,+\,{{a}^{-1}}\,{{r}^{-n+1}}\,\,$

$=\,\,\frac{a}{{{a}^{2}}\,{{r}^{n-1}}}\,\left( a{{r}^{n-1}}\,+\,a{{r}^{n-2}}\,+\,…\,+\,ar\,+\,a \right)$

$=\,\,\frac{1}{{{a}^{2}}{{r}^{n-1}}}\,\times \,S$

$\therefore \,\,\,\,\frac{S}{R}\,\,=\,\,{{a}^{2}}\,{{r}^{n-1}}$

$\therefore \,\,{{\left( \frac{S}{R} \right)}^{n}}\,\,=\,\,{{a}^{2n}}{{r}^{n(n-1)}}$ 

 $\,\therefore \,\,\,\,{{\left( \frac{S}{R} \right)}^{n}}\,\,=\,\,{{P}^{2}}$

Standard 11
Mathematics

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