A geometric series with common ratio $r = -0.9$ has sum to infinity 10000 For this series, (a) find the first term, (b) find the fifth term, (c) find the sum of the first twelve terms, giving this answer to the nearest integer. - Edexcel - A-Level Maths Pure - Question 8 - 2018 - Paper 4

To determine the fifth term of the geometric series, we use the formula for the $n^{th}$ term, which is:

$T_n = a r^{n-1}$

For the fifth term ($n = 5$), we know:

• $a = 19000$
• $r = -0.9$

Substituting these values in:

$T_5 = 19000 \times (-0.9)^{5-1}$ $T_5 = 19000 \times (-0.9)^4$ $T_5 = 19000 \times 0.6561$ $T_5 = 12466.9$

The fifth term is approximately $12467$ (to the nearest integer).

To find the sum of the first twelve terms of the geometric series, we use the formula:

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

where $n = 12$. Substituting the known values:

• $a = 19000$
• $r = -0.9$
• $n = 12$

We calculate:

$S_{12} = 19000 \frac{1 - (-0.9)^{12}}{1 - (-0.9)}$

Calculating the denominator: $S_{12} = 19000 \frac{1 - (-0.9)^{12}}{1 + 0.9}$ $S_{12} = 19000 \frac{1 - 0.282429536481}{1.9}$ $S_{12} = 19000 \frac{0.717570463519}{1.9}$ $S_{12} \approx 19000 \times 0.37765 \approx 7176.19$

Rounding to the nearest integer, the sum of the first twelve terms is $7176$.