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Question

Let a1,a2a3n be an arithmetic progression with a1=3 and a2=7. If a1+a2++a3n=1830, then what is the smallest positive integer m such that m(a1+a2++an)>1830?


A

8

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B

9

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C

10

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D

11

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Solution

The correct option is B

9


a1=3a2=7
common difference =d=73=4a3n=a1+(3n1)×4=3+12n4=12n13+7+11++(12n1)=18303n[6+(3n1)×4]2=18303n[2+12n]=3660n[1+6n]=6106n2+n610=0n=10
Therefore, a10=3+9×4=39a1+a2++a10=3+7++39=5×42=210m×210>1830m>8
Hence smallest integral value of m = 9.


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