Sequences and series (1 Viewer)

[AFGHAN22]

1. Two men A and B commence work on an annual salary of $10000. A receives an increment of $1000 p.a. whilst B receives an increment of $500 each 6 months. Who has received the greater amount after 10 years, and by how much?

Answer = B by $2500 ( Note B’s pay rises last ½ year each)

2. A’s present salary is $390 per week and each year is to receive a rise of $60 per week. B starts with the same salary but receives a half-yearly rise of $30 per week.
Taking a year as exactly 52 weeks, find the total amounts received by each during the first n years.

I only need the amount received by B as I have already obtained B.

Answer = A: $1560n (n+12), B: $780n(2n+25)

thankx in advance guys

 

[damo676767]

i tried to post the answer this morning but it scrwed up, i'l try again

1) forget the bace wage, it isun important

A = 0 + 1000 + 2000 + ... + 9000
= 9000 * 10 / 2
= 45000

B = (0 + 500 + 1000 + ..... + 9500)/2
= 9500 * 20 / 2*2
= 47500

B - A = 2500

2)
A = 390 * 52n + (0 + 60 + 120 + ... + 60*(n-1))*52
= 20280n + 60*(n-1) * n /2 * 52
= 20280n + 1560n(n-1)
= 1560n (13 + n - 1)
= 1560n(n + 12)


B = 390 * 52n + (0 + 30 + 60 + ... + 30*(2n-1)) * 26
= 20280n + 30*(2n-1) * 2n / 2 * 26
= 20280n + 780n(2n - 1)
= 780n (26 + 2n -1)
= 780n (2n + 25)


Any questions?