Question

A geometric series has first term 'a' and common ratio '$r$'. The second term of the series is 4 and the sum to infinity of the series is 25.
If r is taken the larger of its two possible values then the smallest value of n for which $S_n$ exceeds 24, is

• A. 12
• B. 13
• C. 14
• D. 15

Answer 1

Answer: (D)

$\text { if } r =\frac{4}{5} \Rightarrow a =5 $
$ S _{ n }=\frac{ a \left(1- r ^{ n }\right)}{(1- r )}$
$ \frac{5\left(1-\left(\frac{4}{5}\right)^{ n }\right)}{\left(1-\frac{4}{5}\right)}>24 \Rightarrow 25\left(1-\left(\frac{4}{5}\right)^{ n }\right)>24 \Rightarrow \left(\frac{4}{5}\right)^{ n }<\frac{1}{25} $
$\Rightarrow n \log (0.8)<-\log 25 \Rightarrow n \log \frac{5}{4}>2 \log 5$
$\Rightarrow n >\frac{2 \log 5}{\log 5-\log 4} n >\frac{2(1-0.3)}{1-0.3-2 \times 0.3} $
$n >14 .-\cdots$
Smallest value of $n$ is 15