Question 33102
a.  Given sequence is a geometric sequence;
Sum of a finite geometric sequence
S = a1(1-r^n)/(1-r) where a1=first term, r=common ration, n=number of terms;
Here, a1 = 4, r=2, n = 20
Sum = 4(1-2^20)/(1-2) = 4(2^10-1)/1 = 4(1023) = 4092;
...........
b. S4 = 21 
=> 4(a1+a4)/2 = 21 => a1+a4 = 21/2
=> a1+a4 = 21/2;...........(1)
Sum of last three terms = 90
=> 3(a18+a20)/2 = 90
=> a18+a20 = 60 ........ (2)
From (1): a1+[a1+3d] = 21/2 => 2a1+3d = 21/2 ... (3)
From (2): [a1+17d] + [a1+19d] = 90 = > 2a1+36d = 90 =>a1=45-18d;... (4)
Substituting a1 from (4) into (3):
2(45-18d)+3d=21/2 => 90-33d = 21/2 => 33d = 159/2 => d=53/22
From(4): a1=45-18d => a1= 45-477/11 => a1 = 18/11
Answer: First term is 18/11 and the common difference is 53/22;