|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Apr 2007 21:00:32 IST
|
|
|
yaa dats n easy one
let 1st term be a nd common diff. be d
soo we get 6a+69d=225
now by putting the formula s=a/1[2a+(n-1)d] we get
s=24/2[2a+23d]=4{6a+69d}=4*225=900
hence d
|
this reply: 5 points
(with 1 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
