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  • Find a,b and n in the expansion of (a + b)n  if the first three terms of the expansion are 128, 2240, 16800 respectively.Option: 1</str

Find a,b and n in the expansion of (a + b)n  if the first three terms of the expansion are 128, 2240, 16800 respectively.

Option: 1

a=2, b=5 and n=7


Option: 2

a=10, b=2 and n=7


Option: 3

a=5, b=2 and n=5


Option: 4

a=5, b=2 and n=7


Answers (1)

best_answer

The first three terms in the expansion of (a+b)^n

\begin{array}{l}{T_{1}=^{n} C_{0} a^{n-0} b^{0}=a^{n}=128} \\ {T_{2}=^{n} C_{1} a^{n-1} b^{1}=n a^{n-1} b=2240}\\T_{3}=^{n} C_{2} a^{n-2} b^{2}=\frac{n(n-1)}{2} a^{n-2} b^{2}=16800\end{array}

Now, 

\large \begin{array}{l}{\frac{T_2}{T_1}=\frac{n a^{n-1} b}{a^{n}}=\frac{2240}{128}} \\\\ {\frac{n b}{a}=\frac{35}{2}}\end{array} ......(i)

and   

 \large \begin{array}{l}{\frac{T_3}{T_2}=\frac{n(n-1) a^{n-2} b^{2}}{2 n a^{n-1} b}=\frac{16800}{2240}} \\\\ {\frac{(n-1) b}{a}=\frac{30}{2}}\end{array}.......(ii)

Dividing these 2 equations, 

\large \begin{array}{c}{\frac{n}{n-1}=\frac{7}{6}} \\\\ {n=7}\end{array}

Hence, 

T_1=a^7=128 \\\Rightarrow a=(128)^{\frac{1}{7}}=2                  

and

   T_2=n a^{n-1} b=2240 \\\Rightarrow 7\times(2)^6b=2240\\\Rightarrow b=5

hence, option A is correct

 

 

 

Posted by

Shailly goel

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