SOLUTION: how many terms of the A.P. 1,4,7,10,.... are needed to get the sum 715 ?
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Question 884554: how many terms of the A.P. 1,4,7,10,.... are needed to get the sum 715 ?
Answer by skartikey(21) (Show Source): You can put this solution on YOUR website!
s=n/2{2a+(n-1)d}
715=n/2{2X1+(n-1)3}
1430=n(2+3n-3)
3n^2-n-1430=0
3n^2-(66-65)n-1430=0
3n(n-22)-65(n-22)=0
(n-22)(3n-65)=0
=>n-22=0
n=22
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