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જો $\binom{n-1}{4} , \binom{n-1}{5} ,\binom{n-1}{6}$ સમાંતર શ્રેણી હોય તો $n$ શોધો
$15$ અથવા $8$
$10$ અથવા $5$
$15$ અથવા $10$
$8$ અથવા $10$
Solution
$\left( {\begin{array}{*{20}{c}}
{n – 1} \\
4
\end{array}} \right)\,\, + \,\,\left( {\begin{array}{*{20}{c}}
{n – 1} \\
6
\end{array}} \right)\,\, = \,\,2\,\left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right)\,$
$\left( {\begin{array}{*{20}{c}}
{n – 1} \\
4
\end{array}} \right)\,\, + \,\,\left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right)\,\, + \,\left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right)\,\,\, + \,\,\left( {\begin{array}{*{20}{c}}
{n – 1} \\
6
\end{array}} \right)\,\,\, = \,\,4\,\left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right)\,\,\,\,\,\,$
$\left( {_5^n} \right)\, + \,\,\left( {_6^n} \right)\,\, = \,\,4\,\left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right)\,$
$\,\left( {\begin{array}{*{20}{c}}
{n + 1} \\
6
\end{array}} \right)\,\,\, = \,4\,\left( {\begin{array}{*{20}{c}}
{n – 1} \\
5
\end{array}} \right)\,\,\,\,$
$\frac{{n\,(n + 1)(n – 1)(n – 2)(n – 3)(n – 4)}}{{720}}$
$\, = \,\,\frac{{4(n – 1)(n – 2)(n – 3)(n – 4)(n – 5)}}{{120}}$
$n^2 + n = 24(n – 5) $
$n^2 – 23n + 120 = 0$
$ n = 15 $ અથવા $ 8$