find the value of 7.8 bar +7.88 bar +7.888 bar + 7.8888 bar + ….. up to 2014 times (interns of non terminating recurring decimal form
The Correct Answer and Explanation is:
The given series is:7.8‾+7.88‾+7.888‾+7.8888‾+⋯7.\overline{8} + 7.\overline{88} + 7.\overline{888} + 7.\overline{8888} + \cdots7.8+7.88+7.888+7.8888+⋯
Each term in the series is of the form 7.8…‾7.\overline{8 \ldots}7.8… with an increasing number of 8’s in the repeating decimal.
To solve this, we break it down into the form of a geometric series. The general term for this type of recurring decimal can be written as:Sn=7+89⋅10−nS_n = 7 + \frac{8}{9} \cdot 10^{-n}Sn=7+98⋅10−n
Where:
- SnS_nSn represents the nnn-th term in the series.
- The denominator 9 is used because of the repeating decimal (as 0.\overline{8} = 8/9).
- The term 10−n10^{-n}10−n is used to account for the position of the decimal point shifting with each subsequent term.
Step-by-Step Calculation:
- First term (7.\overline{8}): 7.8‾=7+897.\overline{8} = 7 + \frac{8}{9}7.8=7+98 which is the sum of a constant and the fraction from the recurring decimal.
- Second term (7.\overline{88}): 7.88‾=7+88997.\overline{88} = 7 + \frac{88}{99}7.88=7+9988 This term has two repeating digits, so the denominator becomes 99.
- Third term (7.\overline{888}): 7.888‾=7+8889997.\overline{888} = 7 + \frac{888}{999}7.888=7+999888 And so on, as the number of repeating digits increases.
The series can be expressed as the sum:∑n=12014(7+8⋯89⋯9)\sum_{n=1}^{2014} \left( 7 + \frac{8 \cdots 8}{9 \cdots 9} \right)n=1∑2014(7+9⋯98⋯8)
where the number of digits in the numerator and denominator increases as we progress.
Summing the Infinite Series:
To calculate the value of this sum up to 2014 terms, we treat it as an infinite geometric series. The formula for the sum of an infinite geometric series is:S=a1−rS = \frac{a}{1 – r}S=1−ra
Where:
- aaa is the first term,
- rrr is the common ratio between successive terms.
Since the series has an increasing number of 8’s in each term, it becomes more complicated. However, it can be approximated with the following formula:S≈2014⋅(7+89) (approximate)S \approx 2014 \cdot \left( 7 + \frac{8}{9} \right) \text{ (approximate)}S≈2014⋅(7+98) (approximate)
This simplifies to:S≈2014⋅(7.8888)S \approx 2014 \cdot \left( 7.8888 \right)S≈2014⋅(7.8888)
Finally, the total sum up to 2014 terms is:S≈15859.51S \approx 15859.51S≈15859.51
This is the approximate value of the sum of the given series up to 2014 terms.
