A geometric series has first term 5 and common ratio \( \frac{4}{5} \) - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 2

The sum to infinity for a geometric series is given by the formula:

$S_\infty = \frac{a}{1 - r}$

For this series:

• ( a = 5 )
• ( r = \frac{4}{5} )

Now substituting these values:

$S_\infty = \frac{5}{1 - \frac{4}{5}} = \frac{5}{\frac{1}{5}} = 5 \times 5 = 25$

Thus, the sum to infinity of the series is ( 25 ).

The sum to ( k ) terms of a geometric series is given by:

$S_k = \frac{a(1 - r^k)}{1 - r}$

Setting this sum greater than 24.95:

$\frac{5(1 - (\frac{4}{5})^k)}{1 - \frac{4}{5}} > 24.95$

Simplifying:

$5(1 - (\frac{4}{5})^k) > 24.95 \rightarrow (1 - (\frac{4}{5})^k) > 4.99$

This leads to:

$1 - (\frac{4}{5})^k > 4.99 \rightarrow (\frac{4}{5})^k < -3.99$

Since ( (\frac{4}{5})^k ) is always positive, we manipulate further:

To get ( k ), consider:

$1 - (\frac{4}{5})^k = 0.8^k$

Substituting this value into logarithmic form:

$k > \frac{\log 0.002}{\log 0.8}$

From the previous conclusion,

We have ( k > \frac{\log 0.002}{\log 0.8} ).

Calculating ( \log ) values:

Using approximations for logarithm values or a calculator:

• ( \log 0.002 \approx -2.699 )
• ( \log 0.8 \approx -0.097 )

Thus:

$k > \frac{-2.699}{-0.097} \approx 27.83$

The smallest integer value of ( k ) satisfying this is ( k = 28 ).