Can anybody help me in this​

a) I donot agree with him because as we see the sequence nicely we will find an interval of 8 between any two terms and 32 and 33 don't match them

Step-by-step explanation:

b) and c) Solution:

Given,

First term (a) = 4

Common difference (d) = 12-4 = 8

Now,

Since, they are in A.P.

nth term = a + (n-1)d

              = 4 + 8(n-1)

              = 4 + 8n-8

              = 8n-4

Again,

1st method for c)

100th term = a + 99d = 4 + 99*8 = 4 + 792 = 796

200th term = a + 199d = 4 + 199*8 = 4 + 1592 = 1596

2nd method for c)

100th term = 8*100 - 4 = 800 - 4 = 796

200th term = 8*200 - 4 = 1600 - 4 = 1596

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p