. Find the sum of the given arithmetic series.
The sequence whose general term is an=3n represents the positive multiples of 3. Find the sum of the first 102 positive multiples of 3.
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Sum = (n/2)( 2a +(n-1)d )
just about remembering formulas
I dont even need to know what d is ( I could get it by simultaneously eqns ) , because remember that general Tn ( or in your case "an" ) = a +(n-1)d
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so the part inside the bracket is:
a+ ( a+(n-1)d )
now I will call "an" T(n) , ie T(n) = 3n
so our sum becomes:
Sum = (n/2) ( a +T(n) )
Now n= 102 ( because we want first 102 terms ) and T(n) = 3n ( thats given to us )
so sum = (n/2) ( a+ 3n )
now we just need to know what "a" is , we already know "n" from before ( its 102 )
so if we sub n=1 into our T(n) formula that will give us the first term ( which is "a" )
so T(1) = a = 3(1) = 3
now sub it in
sum = (102/2) ( 3 + 3(102) ) = 51 ( 309) = 15759
wow i dont know how this comes so easy for u, thank you. i have like 3 more problems ;(