Question

Please solve q 11 I will surely mark you brailiest

Answer 1

Solution :

$\frac{ {6}^{n + 3} - 32. {6}^{n + 1} }{ {6}^{n + 2} - 2. {6}^{n + 1} } \\ \\ = \frac{ {6}^{(n + 1 )+ 2} - 32. {6}^{n + 1} }{ {6}^{(n + 1) + 1} - 2. {6}^{n + 1} } \\ \\ = \frac{ {6}^{n + 1} \times 6^2 - 32. {6}^{n + 1} }{ {6}^{n + 1} \times6^1 - 2.{6}^{n + 1} }\\ \\ = \frac{ {6}^{n + 1}( {6}^{2} - 32) }{ {6}^{n + 1}( {6}^{1} - 2) } \\ \\ = \frac{36 - 32}{6 - 2} = \frac{4}{4} = \mathbf{1}$

Hence ,

$\boxed{ \textbf{Ans : (d) 1}}$

Answer 2

Hey Prave,

Here is your solution.

6^( n + 3 )- 32  × 6^( n + 1 )
=  --------------------------------
   6^ ( n + 2 ) - 2 × 6^ ( n + 1 ) 

By taking 6^( n + 1 ) as common in numerator and denominator.

  6^ ( n + 1 ) { 6^ ( n + 3 - n - 1 )  - 32 }
= ------------------------------------------
   6^ ( n + 1 ) { 6^ ( n + 2 - n - 1 ) - 2 }

{ 6^( 2 ) - 32 }
=  ---------------------
      { 6^ ( 1 ) - 2 }

    36 - 32
= --------------
     6 - 2

= 4 ÷ 4

= 1

The answer is option ( d ) 1.