- Home
- Standard 11
- Mathematics
જો $\left( {_{\,\,\,4}^{n - 1}} \right),{\text{ }}\left( {_{\,\,\,5}^{n - 1}} \right)\,$ અને $\left( {_{\,\,\,6}^{n - 1}} \right)\,$ સમાંતર શ્રેણીમાં હોય તો................. મળે
$\left( {_{\,\,\,6}^{n + 1}} \right) = 2\left( {_{\,\,\,5}^{n - 1}} \right)$
$2\left( {_{\,\,\,\,6}^{n + 1}} \right) = \left( {_{\,\,\,5}^{n - 1}} \right)$
$\left( {_{\,\,\,\,6}^{n + 1}} \right) = 4\left( {_{\,\,\,\,5}^{n - 1}} \right)$
$4\left( {_{\,\,\,\,6}^{n + 1}} \right) = \left( {_{\,\,\,5}^{n - 1}} \right)$
Solution
આપેલ પદ સમાંતર શ્રેણી માં હોવાથી
$ \left( {\begin{array}{*{20}{c}}
{n – 1} \\
4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
{n – 1} \\
6
\end{array}} \right) = 2\left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right)\,\,;\,\,$
બંને બાજુ $2\,\left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right)\,$ ઉમેરતા
$\left( {\begin{array}{*{20}{c}}
{n – 1} \\
4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
{n – 1} \\
6
\end{array}} \right) = 4\left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right)\,\,$
$\,\,\left( {\begin{array}{*{20}{c}}
n \\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
n \\
6
\end{array}} \right) = 4\left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right)\,$
$\left( {\begin{array}{*{20}{c}}
{n + 1} \\
6
\end{array}} \right) = 4\left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right)$ મળે
