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Ok, now I get it!ill explain u how (n-2)d=(n-1)d-d
we expand (n-2)d : and get nd-2d
we change equation to nd-d-d. they both are equal its just that i changed -2d into -d-d
then we factor d outside and get : (n-1)d-d
hope that helped
if still not ask again by stating what part u didnt get.
What about this part? "WE SUBSTITUTE THE VALUE OF z+(n-1)d
THUS WE GET: S=n/2[z+{(y+d)/2}"
I subsitiuted it...but I couldnt come up with --> S=n/2[z+{(y+d)/2} Could you pls explain?