How-many-terms-in-the-geometric-progression-1-1-1-1-21-1-331-will-be-needed-so-that-the-sum-of-the-first-n-terms-is-greather-than-20-

Question Number 99603 by bemath last updated on 22/Jun/20

How many terms in the geometric

progression 1, 1.1, 1.21, 1.331, \dots

will be needed so that the sum of

the first n terms is greather than 20 ?

Commented by john santu last updated on 22/Jun/20

The sequence is a geometric progression

with {\begin{cases} a = 1 \\ r = 1.1 \end{cases}

we want to find the smallest value of

n such that S_{n} > 20

\Rightarrow S_{n} = \frac{a (1 - r^{n})}{1 - r} = \frac{1 . (1 - {(1.1)}^{n}}{1 - 1.1} > 20

\frac{1 - {(1.1)}^{n}}{- 0.1} > 20; {(1.1)}^{n} - 1 > 2

{(1.1)}^{n} > 3; n > \frac{\ln 3}{\ln 1.1}; n > 11.526

and therefore the smallest number

value of n is 12

Commented by bemath last updated on 22/Jun/20

thank you both

Answered by Rio Michael last updated on 22/Jun/20

S_{n} = \frac{a (r^{n} - 1)}{r - 1}

S_{n} = \frac{{(1.1)}^{n} - 1}{1.1 - 1} = \frac{{(1.1)}^{n} - 1}{0.1}

for S_{n} > 20, \Rightarrow \frac{{(1.1)}^{n} - 1}{0.1} > 20 \Rightarrow {(1.1)}^{n} > 3

n \log (1.1) > \log 3

\Rightarrow n > 11.52, but n \in Z \Rightarrow n = 12 .