Question

The sum to infinity of the progression \[9-3+1-\dfrac{1}{3}+\dfrac{1}{9}-....\] is
A. \[\dfrac{27}{4}\]
B. \[\dfrac{9}{2}\]
C. \[9\]
D. \[\dfrac{15}{2}\]

Answer 1

Hint: The question is formed by values which are forming a pattern in an infinite series with \[9\] as the first number and the difference between the numbers are \[\dfrac{-1}{3}\] . We will use the infinite series pattern formula from geometric progression where sum of an infinite series is given as:
Sum of the series \[=\dfrac{a}{1-r}\]
where \[a\] is the first digit and \[r\] is the gap pattern between the numbers.

Complete step-by-step answer:
Geometric series is a series in which the first element is found by multiplying it with the difference of the series.
 Now as we can see in the question the first number in the series is given as \[a=9\] .
And to find the difference between the number of the given pattern as \[r=-\dfrac{1}{3}\] .
Now that we have the value of \[a\] and \r\, we can find the value of sum of the series as:
Sum of the series
\[=\dfrac{a}{1-r}\]
\[=\dfrac{9}{1-\left( -\dfrac{1}{3} \right)}\]
\[=\dfrac{9\times 3}{4}\]
\[=\dfrac{27}{4}\]
Therefore, the sum of the series is given as \[\dfrac{27}{4}\] .
So, the correct answer is “ \[\dfrac{27}{4}\] ”.

Note: Student may go wrong if they use the finite formula of geometric progression to find the sum of the progression as the formula for the finite series sum is: \[S=a\dfrac{\left( {{r}^{n}}-1 \right)}{r-1}\] .