The common difference of an arithmetic sequence is 6 and its one te is 25. Can the sum of any 15 terms in this sequence be 2018? Why?
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Sum of n terms in this A.P. is
S=n/2(2a+(n-1)d)=2018
15/2(2a+(15–1)6)=2018
2a=((2018×2)/15)–84
a=2776/30
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