Answer:
3040
Step-by-step explanation:
given arithmetic progression is
70,100,130,...
here
first term (a)=70
common difference (d)=100-70=30
number of term n=100
using the formula of arithmetic progression
an=a+(n-1)d
a100=70+(100-1)30
a100=70+99×30
a100=70+2970
a100=3040
Lets round it to the nearest ten
A 97 ====> 100
B 118 ===> 120
C 179 ===> 180
D 5091 ==> 5090
No result yet, lets round to the nearest hindred.
A 97 ====> 100
B 118 ===> 100
C 179 ===> 180
D 5091 ==> 5100
As we can see only A give the same result when we round it to the nearest hundred and nearest ten.
So a=2 since a is next to the first 2 it would be 2x2 which equals 4 next it’s B and B= -3 so -3x4 equals -12 and then c also equals 2 so it would be -12x2 and that equals -24. Moving on to the next part as I said a=2 so it would be 3x2 and that equals 6 next it’s B again so 6x -3 equals -18 so now take -24 and -18 and subtract them and you would get -6. Sorry if this answer is wrong I did my best...hope it helped.
See attached PDF. (The censor thinks there are some unseemly words in there.)
