SOLUTION: find x in the series 1+4+7+...+x=715

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Question 1096099: find x in the series 1+4+7+...+x=715
Answer by Boreal(15235) About Me  (Show Source):
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Sum=(n/2)(a1+an); also an=a1+d(n-1)=1+3n-3=3n-2
715=(n/2)(a1+an)
1430=n(a1+an); a1=1
1430=n*(an+1); substitute 3n-2 for an
1430=n(3n-1)=3n^2-n
3n^2-n-1430=0
n=(1/6) 1 +/- sqrt(1+17160); sqrt(17161)=131
n=+ (positive root only) (1/6)*132=22
Therefore, 715=11(1+an)
65=a1+an=1+an
an=x=64 ANSWER
The sum of 1+4+7+10+13+16+19+22+25+28+31+34+37+40+43+46+49+52+55+58+61+64=715