Find the sum of the frist 40 positive intergers divisible by 6
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wk that the first 40 positive integers divisible by 6 are 6,12 ,18 ,24 .....
this is an AP with a=6 and d=6
Sn= n/2 [2a+(n-1)d]
S40 =20[12+(40-1 )d]
=>20[12+234]= 4920
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