Question

a man accepts a position with an initial salary of money $1800 per year if his salary increases at the end of every year by $1500 what would be his annual salary at Te begging of the 11th year

Answer 1

1800 + 11*1500

Answer 2

I dntknw yeerrrr

Answer 3

33000

Answer 4

31,500

Answer 5

A1=18,000.00 , d=1,500.00 . The beginning of the 11^th year is the same as the end of the 10^th year. ->Thus, A10 = A1 + 9d = 18,000.00 + 9 * 1,500.00 = 31,500.00 . Hence , his annual salary at beginning of 11^th year is $31,500.

Answer 6

if (gn) is a geometric sequence with gn>0 for all n an element of N, then prove that (ln gn) is an arithmetic sequence.

Answer 7

A1=a+1500 so that A1=19500 b/c a=( and also d=1500.we find a10th b/c it's at the beginning of 11th year. The formula is
an=A1+d (n-1) so that a10=19500+1500(10-1).
=19500+1500(9)
=19500+13500
=33,000