$5/11, 5/121, 5/1331, 5/14641, ...$ to $n$ terms.
Finding the Sum of the Geometric Progression
The given series is a Geometric Progression (G.P.): $5/11, 5/121, 5/1331, 5/14641, \dots$
Identify G.P. Parameters
- The first term, $a$, is $5/11$.
- To find the common ratio, $r$, divide the second term by the first term: $r = \frac{5/121}{5/11} = \frac{5}{121} \times \frac{11}{5} = \frac{11}{121} = \frac{1}{11}$
- The number of terms is $n$.
Apply the Sum Formula
The formula for the sum of the first $n$ terms of a G.P. where $|r| < 1$ is:
$ S_n = \frac{a(1-r^n)}{1-r} $Calculate the Sum
Substitute the values of $a$ and $r$ into the formula:
$ S_n = \frac{\frac{5}{11} \left(1 - \left(\frac{1}{11}\right)^n\right)}{1 - \frac{1}{11}} $First, calculate the denominator:
$ 1 - \frac{1}{11} = \frac{11}{11} - \frac{1}{11} = \frac{10}{11} $Now substitute this back into the sum formula:
$ S_n = \frac{\frac{5}{11} \left(1 - \left(\frac{1}{11}\right)^n\right)}{\frac{10}{11}} $Simplify the expression:
$ S_n = \frac{5}{11} \times \frac{11}{10} \times \left(1 - \left(\frac{1}{11}\right)^n\right) $ $ S_n = \frac{5}{10} \times \left(1 - \left(\frac{1}{11}\right)^n\right) $ $ S_n = \frac{1}{2} \left(1 - \left(\frac{1}{11}\right)^n\right) $Conclusion
The sum of the G.P. to $n$ terms is $\frac{1}{2}(1-(\frac{1}{11})^n)$.
Important Questions from Progression (Notes)
- The sum of 16 terms of the series $\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + .....$ is :
- An auditorium has 8 seats in the first row, with every row to follow having 4 more seats than its preceding row. The total capacity is 416. What is the minimum number of rows needed to seat 150 people?
- The $5^{\text{th}}$ and $9^{\text{th}}$ terms of an arithmetic progression are 7 and 13 respectively. What is the $15^{\text{th}}$ term?
Suppose $a_1, a_2,..., a_{300}$ are integers such that $a_{i-1}+ a_i+ a_{i+1} = 2025$ for all $i = 2,3, ..., 299$.
If $a_7 = -5, a_9 = 37$, then the value of $a_{106}$ isThe value of $1 + (\frac{1}{2^1} + \frac{1}{3}) + (\frac{1}{2^2} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}) + ... + (\frac{1}{2^9} + ... + \frac{1}{1023})$ lies between
