The first term of a geometric series is 120 - Edexcel - A-Level Maths Pure - Question 5 - 2006 - Paper 2

The nth term of a geometric series is given by:

$u_n = ar^{n-1}$

First, we calculate the 5th term ( u_5 ) and the 6th term ( u_6 ):

• For ( u_5 ):

$u_5 = 120 \left(\frac{3}{4}\right)^{5-1} = 120 \left(\frac{3}{4}\right)^{4} \approx 120 \times 0.3164 \approx 37.9685$

• For ( u_6 ):

$u_6 = 120 \left(\frac{3}{4}\right)^{6-1} = 120 \left(\frac{3}{4}\right)^{5} \approx 120 \times 0.2373 \approx 28.4766$

The difference between the 5th and 6th terms is:

$u_5 - u_6 \approx 37.9685 - 28.4766 = 9.4919 \approx 9.49$

The sum of the first n terms of a geometric series is given by:

$S_n = a \frac{1 - r^n}{1 - r}$

For the first 7 terms, we substitute:

• ( a = 120 )
• ( r = \frac{3}{4} )
• ( n = 7 )

Calculating:

$S_7 = 120 \frac{1 - \left(\frac{3}{4}\right)^7}{1 - \frac{3}{4}} = 120 \frac{1 - \left(\frac{3}{4}\right)^7}{\frac{1}{4}} = 120 \cdot 4 \left( 1 - \left(\frac{3}{4}\right)^7 \right)$

Calculating ( \left(\frac{3}{4}\right)^7 \approx 0.1335 ):

$S_7 \approx 480 \left( 1 - 0.1335 \right) = 480 \times 0.8665 \approx 416.0$

We know from the problem statement that the sum of the first n terms is greater than 300:

$S_n > 300$

Using the formula for the sum of the first n terms:

$S_n = a \frac{1 - r^n}{1 - r} > 300$

Substituting the known values:

$120 \frac{1 - \left(\frac{3}{4}\right)^n}{1 - \frac{3}{4}} > 300$

This simplifies to:

$480(1 - (\frac{3}{4})^n) > 300$

Rearranging:

$1 - \left(\frac{3}{4}\right)^n > \frac{300}{480} = \frac{5}{8}$

So:

$\left(\frac{3}{4}\right)^n < \frac{3}{8}$

Taking logarithm on both sides:

$n \cdot \log(\frac{3}{4}) < \log(\frac{3}{8})$

Substituting values: $n > \frac{\log(\frac{3}{8})}{\log(\frac{3}{4})} \approx \frac{-0.1249}{-0.1249} \approx 4$

Thus, the smallest possible value of n is ( n = 4 ).