The first term of a geometric series is 120 - Edexcel - A-Level Maths Pure - Question 6 - 2006 - Paper 2

The nth term of a geometric series can be calculated using:

$u_n = a \cdot r^{n-1}$

Where ( a = 120 ) and ( r = \frac{3}{4} ).

Calculating the 5th term:

$u_5 = 120 \cdot \left( \frac{3}{4} \right)^{5-1} = 120 \cdot \left( \frac{3}{4} \right)^4 = 120 \cdot \frac{81}{256} \approx 38.76$

Calculating the 6th term:

$u_6 = 120 \cdot \left( \frac{3}{4} \right)^{6-1} = 120 \cdot \left( \frac{3}{4} \right)^5 = 120 \cdot \frac{243}{1024} \approx 28.47$

Now, finding the difference:

$u_5 - u_6 \approx 38.76 - 28.47 = 10.29$

So the difference is approximately 10.29.

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:

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

Calculating it:

1. Substitute values: [ S_7 = 120 \cdot 4 \cdot \left( 1 - \left( \frac{3}{4} \right)^7 \right) ]

2. Simplifying the term ( \left( \frac{3}{4} \right)^7 ): [ \left( \frac{3}{4} \right)^7 \approx 0.1335 ]

3. Continuation: [ S_7 = 480 \cdot \left( 1 - 0.1335 \right) = 480 \cdot 0.8665 \approx 416 $$

We need to find the smallest ( n ) such that:

$S_n > 300$

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

$S_n = 120 \frac{1 - \left( \frac{3}{4} \right)^n}{1 - \frac{3}{4}} = 480 (1 - \left( \frac{3}{4} \right)^n)$

Setting up the inequality:

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

1. Simplifying the expression: [ 1 - \left( \frac{3}{4} \right)^n > \frac{300}{480} = \frac{5}{8} ]

2. Rearranging: [ \left( \frac{3}{4} \right)^n < \frac{3}{8} ]

3. Taking logarithms on both sides: [ n \log \left( \frac{3}{4} \right) < \log \left( \frac{3}{8} \right) ]

4. Solving for ( n ): [ n > \frac{\log \left( \frac{3}{8} \right)}{\log \left( \frac{3}{4} \right)} ] [ n > 4 ] (approximately) Thus, the smallest value of ( n ) is 4.