Answer :
S= a(1-r ⁿ) /(1-r) & when r is < 1 & n →∞, the sum becomes S= a/(1-r)
Progression P is 2,1, 1/2,1/4,.. ==> r = 1/2
Sum of P when n--> ∞ = 2/(1-1/2) ==> S =4
Progression Q is 3,1,1/3,1/9,....==> r = 1/3
Sum of Q when n--> ∞ = 3/(1-1/3) ==> S =4.5
GIVEN THAT 4, 9/2 & X (to be calculated later is a geometric Progression, hence 9/2 - 4 = 0.5 =d (common difference , so X = 4.5+0.5 & X = 5
Sum of R =5 Then => 5= 4 /(1-r) & r=1/5
Then the sum of R = 4/(1-1/5) ==> S of R =5
Progression P is 2,1, 1/2,1/4,.. ==> r = 1/2
Sum of P when n--> ∞ = 2/(1-1/2) ==> S =4
Progression Q is 3,1,1/3,1/9,....==> r = 1/3
Sum of Q when n--> ∞ = 3/(1-1/3) ==> S =4.5
GIVEN THAT 4, 9/2 & X (to be calculated later is a geometric Progression, hence 9/2 - 4 = 0.5 =d (common difference , so X = 4.5+0.5 & X = 5
Sum of R =5 Then => 5= 4 /(1-r) & r=1/5
Then the sum of R = 4/(1-1/5) ==> S of R =5