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The sum of n terms of the sequence 3,9,15,21 is 7500.Determine the value of n.

2 answers

Sn=½n(2a+(n-1)d)
Or
Sn=½n[(a+(a+(n-1)d]

Remember that
Tn=a+(n-1)d

a=3 d=6

Tn=3+(n-1)6
Tn=3+6n-6
Tn=(6n-3)

But sn=½n(a+tn)=7500

½n(3+6n-3)=7500

½n(6n)=7500

3n²=7500

n²=2500

n²=50²

n=50

it is correct

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