Answer to Question #323498 in Algebra for ild

Question

An engineer earns £21016 per annum and receives annual increments of 4%. Determine the salary in the 9th year and calculate the total earnings in the first 11 years.

Answer 1

$b_1=21016\\b_n=1.04b_{n-1}\\b_n=b_1q^{n-1}\\b_9=21016\cdot 1.04^8=28761.8\\b_1+b_2+...+b_{11}=\frac{b_1\left( q^{11}-1 \right)}{q-1}=21016\frac{1.04^{11}-1}{1.04-1}=283429$