Question

please answer the question
0.6666...+1.262626......+2.3535​

Answer 1

Answer:

Required value of 0.6666.... + 1.262626.... + 1.353535.... is 424 / 99.

Step-by-step explanation:

Let the algebraic value of 0.6666.... be a, 0.6666.... can also be written as 0.$\bar{6}$, since 6 is repeating.

= > a = 0.$\bar{6}$

Multiply both sides by 10, as there is only one number under bar :

= > 10 x a = 10 x 0.$\bar{6}$

= > 10a = 6.$\bar{6}$

Subtract a from the above equation :

10a = 6.$\bar{6}$

- a = - 0.$\bar{6}$

9 a = 6

= > a = 6 / 9

= > a = 2 / 3

Similarly,

Let the algebraic value of 1.262626.... be b, 1.261626.... can also be written as 1.$\bar{26}$, since 26 is repeating.

= > b = 1.$\bar{26}$

Multiply both sides by 100, as there are two numbers under bar :

= > 100 x b = 100 x 1.$\bar{26}$

= > 100b = 126.$\bar{26}$

Subtract b from the above equation :

100b = 126.$\bar{26}$

- b = - 1.$\bar{26}$

99 b = 125

= > b = 125 / 99

Similarly,

Let the algebraic value of 2.353535.... be c, 2.353535.... can also be written as 0

2.$\bar{35}$

= > c = 2.$\bar{35}$

Multiply both sides by 100, as there are two numbers under bar :

= > 100 x c = 100 x 2.$\bar{35}$

= > 100c = 235.$\bar{35}$

Subtract a from the above equation :

100c = 235.$\bar{35}$

- c = - 2.$\bar{35}$

99c = 233

= > c = 233 / 99

Thus,

= > 0.6666.... + 1.262626.... + 1.353535....

= > a + b + c

From above,

= > 2 / 3 + 125 / 99 + 233 / 99

= > ( 66 + 125 + 233 ) / 99

= > 424 / 99

Hence the required value of 0.6666.... + 1.262626.... + 1.353535.... is 424 / 99.