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7.Binomial Theorem
medium
જો ${(1 + x)^n}$ ના વિસ્તરણમાં $2^{nd}$, $3^{rd}$ અને $4^{th}$ પદના સહગુણક સમાંતર શ્રેણી માં હોય તો ${n^2} - 9n$ = . . . .
A
$-7$
B
$7$
C
$14$
D
$-14$
Solution
(d) Coefficients of $2^{nd}$, $3^{rd}$ and $4^{th}$ terms are respectively $^n{C_1},{\,^n}{C_2}$ and $^n{C_3}$ are in $A.P.$
==> ${2.^n}{C_2} = {\,^n}{C_1} + {\,^n}{C_3}$
==> $\frac{{2n!}}{{2\,!\left( {n – 2} \right)\,!}} $
= $\frac{{n!}}{{(n – 1)!}} + \frac{{n!}}{{3!\,\left( {n – 3} \right)\,!}}$ On solving, ${n^2} – 9n + 14 = 0\,\,$
$\Rightarrow \,\,{n^2} – 9n = – 14$.
Standard 11
Mathematics
