Question

if 1/3×9 + 1/9×15 +..........upto 1/93×99 , then root 33x equals to​

Answer 1

√33x  = 4/3  if x = 1/3×9 + 1/9×15 +..........upto 1/93×99

Step-by-step explanation:

x = 1/3×9 + 1/9×15 +..........................+ 1/93 * 99

1/3* 9  =  (1/6) ( 1/3  - 1/9)

1/9*15  = (1/6) (1/9 - 1/5)

x  = (1/6)  (   1/3  - 1/9  +  1/9  - 1/15   +...............................+ 1/93 - 1/99)

=> x= (1/6) ( 1/3  - 1/99)

=>  x  = ( 1/6)  ( 33 - 1) /99

=> x = 32 / (6 * 99)

=> x = 16 ( 3 * 99)

=>  33x =  33 * 16 / ( 3 * 99)

=> 33x  = 16 /(3 * 3)

=> 33x  =  (4/3)²

=> √33x  = 4/3

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Answer 2

Given: $\frac{1}{3\times 9}+\frac{1}{9\times 15}+.........+\frac{1}{93\times 99}$

To Find: Find the value of $\sqrt{33x}$.

Step-by-step explanation:

Lets,

$x=\frac{1}{3\times 9}+\frac{1}{9\times 15}+.........+\frac{1}{93\times 99}$

First terms,

$\frac{1}{3\times 9}=\frac{1}{6}(\frac{1}{3}-\frac{1}{9})$

Second terms,

$\frac{1}{9\times 15}=\frac{1}{6}(\frac{1}{9}-\frac{1}{15})$

Similarly all terms will be come.

∴ $x=\frac{1}{6}(\frac{1}{3}-\frac{1}{9})+\frac{1}{6}(\frac{1}{9}-\frac{1}{15})+.............+\frac{1}{6}(\frac{1}{93}-\frac{1}{99})$

Take $\frac{1}{6}$ common from above equation,

⇒ $x=\frac{1}{6}(\frac{1}{3}-\frac{1}{9}+\frac{1}{9}-\frac{1}{15}+.............+\frac{1}{93}-\frac{1}{99})$

⇒$x=\frac{1}{6}(\frac{1}{3}-\frac{1}{99})$

⇒$x=\frac{1}{6}(\frac{33-1}{99})$

⇒$x=\frac{1}{6}\times \frac{32}{99}$

Multiply $33$ on both sides,

⇒$33x=\frac{1}{6}\times \frac{32}{99}\times 33$

⇒$33x=\frac{16}{9}$

⇒$33x=(\frac{4}{3}) ^{2}$

Take root on both sides,

∴ $\sqrt{33x}=\frac{4}{3}$

So, The value of $\sqrt{33x}$ is $\frac{4}{3}$.