Question

find the sum of series for .9,.91, .92, .100 terms

Answer 1

2.83 according to me I hope it helps

Answer 2

let a1 = .9
a2=.91
a3= .92 ,... 100 terms

first term = a= a1 = 0.9
common difference = d= a2-a1 = .91- 0.9 = 0.01
number of term = n= 100
sum of n terms in Ap = Sn = n/2[2a+(n-1)d]

= 100/2 [2*0.9 + (100-1) (0.01)]
= 100/2 [ 1.8 +99*0.01]
= 100/2 (1.8 +0.99)
= (100 *2.79)
=279
s100 = 279