Find the sum of all the multiples of 3 lying between 250 and 1000.
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Answer:
250 to 1000
750 no.
252 is first multiple of 3
999 is last multiple of 3
Tn =end term=a+ {n-1}d
999=252+ (n-1)d
999-252=3n-3
3n=750
n=250
Sn = N/2(a+l)
=125(252+999)
Sn =156,375
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