Answer:
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
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