Question

Find the sum of 0.50+0.52+0.54+... to 40 terms.

Answer 1

$S=0.50+0.52+0.54+...$ to $40$ terms

$a=0.50,d=0.52-0.50=0.02,n=40$

$\therefore S_{40}=\frac{40}{2}[2\times 0.52+(40-1)0.02]$

$\therefore S_{21}=20[1.04+0.78]=20\times 1.82=36.4‬$