Answer to Question #208430 in Algebra for Areeba

Question

The second term of a geometric series is 3 and the common ratio is 4/ 5 . Find the sum of first 23 consecutive terms of this series. 

Answer 1

Consider the geometric series $a, ar, ar^2, ...$

$$a_n=ar^{n-1}, n=1, 2, ...$$

$$S_n=\dfrac{a(1-r^n)}{1-r}, r\not=1$$

Given $r=\dfrac{4}{5}, ar=3.$

Find $a$

$$a(\dfrac{4}{5})=3=>a=\dfrac{15}{4}$$

$$S_{23}=\dfrac{\dfrac{15}{4}(1-(\dfrac{4}{5})^{23})}{1-\dfrac{4}{5}}$$

$$S_{23}=\dfrac{75}{4}\big(1-(\dfrac{4}{5})^{23}\big)$$