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Find A, B and N in the Expansion of (A + B)N, If the First Three Terms in the Expansion Are 729, 7290 and 30375 Respectively.

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प्रश्न

Find a, b and n in the expansion of (a + b)n, if the first three terms in the expansion are 729, 7290 and 30375 respectively.

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उत्तर

\[\text{ We have: } \]

\[ T_1 = 729, T_2 = 7290 \text{ and }  T_3 = 30375\]

\[\text{ Now,}  \]

\[ ^{n}{}{C}_0 a^n b^0 = 729\]

\[ \Rightarrow a^n = 729\]

\[ \Rightarrow a^n = 3^6 \]

\[ ^{n}{}{C}_1 a^{n - 1} b^1 = 7290\]

\[^{n}{}{C}_2 a^{n - 2} b^2 = 30375\]

\[\text{ Also, } \]

\[\frac{^{n}{}{C}_2 a^{n - 2} b^2}{^{n}{}{C}_1 a^{n - 1} b^1} = \frac{30375}{7290}\]

\[ \Rightarrow \frac{n - 1}{2} \times \frac{b}{a} = \frac{25}{6} . . . (i)\]

\[ \Rightarrow \frac{(n - 1)b}{a} = \frac{25}{3}\]

\[\text{ And } , \]

\[\frac{^{n}{}{C}_1 a^{n - 1} b^1}{^{n}{}{C}_0 a^n b^0} = \frac{7290}{729}\]

\[ \Rightarrow \frac{nb}{a} = \frac{10}{1} . . . (ii)\]

\[\text{ On dividing (ii) by (i), we get } \]

\[\frac{\frac{nb}{a}}{\frac{(n - 1)b}{a}} = \frac{10 \times 3}{25}\]

\[ \Rightarrow \frac{n}{n - 1} = \frac{6}{5}\]

\[ \Rightarrow n = 6\]

\[\text{ Since } , a^6 = 3^6 \]

\[\text{ Hence,}  a = 3\]

\[\text{ Now } , \frac{nb}{a} = 10\]

\[ \Rightarrow b = 5\]

 

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अध्याय 18: Binomial Theorem - Exercise 18.2 [पृष्ठ ४०]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 18 Binomial Theorem
Exercise 18.2 | Q 34 | पृष्ठ ४०

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