For-a-sequence-lt-a-n-gt-a-1-2-a-n-1-a-n-1-3-then-r-1-20-a-r-is-equal-to-

Question Number 56411 by gunawan last updated on 16/Mar/19

For a sequence < a_{n} >; a_{1} = 2, \frac{a_{n + 1}}{a_{n}} = \frac{1}{3},

then \underset{r = 1}{\sum^{20}} a_{r} is equal to

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19

\frac{a_{2}}{a_{1}} = \frac{1}{3} = \frac{a_{2}}{2} \to a_{2} = \frac{2}{3}

\frac{a_{3}}{a_{2}} = \frac{1}{3} = \frac{a_{3}}{\frac{2}{3}} \to a_{3} = \frac{2}{3^{2}}

\frac{a_{4}}{a_{3}} = \frac{1}{3} = \frac{a_{4}}{\frac{2}{3^{2}}} \to a_{4} = \frac{2}{3^{3}}

S_{n} = a_{1} + a_{2} + a_{3} + \dots + a_{n}

= \frac{2}{1} + \frac{2}{3} + \frac{2}{3^{2}} + \dots + \frac{2}{3^{n - 1}}

A = 2 r = \frac{1}{3}

S_{n} = \frac{2 (1 - \frac{1}{3^{n}})}{1 - \frac{1}{3}} \to \frac{2 (1 - \frac{1}{3^{n}})}{\frac{2}{3}} \to 3 (1 - \frac{1}{3^{n}})

s o s_{20} = 3 (1 - \frac{1}{3^{20}})