IBDP Maths SL 1.8 The sum of infinite geometric sequences AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
This is an infinite geometric series with
\( a=\dfrac{53}{1000}, \qquad r=\dfrac{1}{100}. \)
Since \( |r|<1 \), the series converges and the sum to infinity is
\( S_{\infty}=\frac{a}{1-r}. \)
Substitute the values:
\( S_{\infty} = \frac{\frac{53}{1000}} {1-\frac{1}{100}} = \frac{\frac{53}{1000}} {\frac{99}{100}} = \frac{53}{990}. \)
This represents the repeating part \(0.0535353\ldots\).
✅ Answer: \( \boxed{S_{\infty}=\dfrac{53}{990}} \)
(b)
The decimal can be written as
\( 0.2\overline{53} = 0.2+0.0535353\ldots \)
Using the result from part (a):
\( 0.2+\frac{53}{990} = \frac{1}{5}+\frac{53}{990}. \)
Convert to a common denominator:
\( \frac{198}{990}+\frac{53}{990} = \frac{251}{990}. \)
The numbers \(251\) and \(990\) have no common factors, so this fraction is already in simplest form.
✅ Answer: \( \boxed{\dfrac{251}{990}} \)