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