$5/11, 5/121, 5/1331, 5/14641, ...$ to $n$ terms.
The given series is a Geometric Progression (G.P.): $5/11, 5/121, 5/1331, 5/14641, \dots$
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} $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) $The sum of the G.P. to $n$ terms is $\frac{1}{2}(1-(\frac{1}{11})^n)$.
If a, b and c are in Geometric Progression and $a^\frac{1}{x} = b^\frac{1}{y} = c^\frac{1}{z}$ then, x, y, z are in ________.
Which of the following statement is true about the geometric series
$ 1 + r +r^2 + r^3 + ...............; (r > 0) $?
$6240$ रुपये की राशि $30$ किस्तों में इस प्रकार चुकाई जाती है कि प्रत्येक किस्त पिछली किस्त से $10$ रुपये अधिक है । पहली किस्त की मूल्य ____________है।