Precalculas
posted by G on .
The sum of the first 30 terms of one arithmetic sequence is 300 more than the sum of the first 30 terms of another arithmetic sequence. what could the Â sequences be?

assume they have the same first term.
Then 29(d2d1) = 300
29 does not divide 300, but it does divide 290
So, in order for s2=s1+300, a2a1=10
one solution would be
a1=0 d1=1
a2=10 d2=11
0+29=29
10+29*11 = 329
a1+29d1 + 300 = (a1+10) + 29(d1+10)