find the sum first 24terms of the list of number whose n term given by a=3+2n
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a1=3+2(1)=5
a2=3+2(2)=7
a3=3+2(3)=9
d=2
Sn=n/2(2a+(n-1)d)
S24 =24/2(2*5+(24-1)2)
=12(10+(23)2)
=12(10+46)
12(56)
=672
a2=3+2(2)=7
a3=3+2(3)=9
d=2
Sn=n/2(2a+(n-1)d)
S24 =24/2(2*5+(24-1)2)
=12(10+(23)2)
=12(10+46)
12(56)
=672
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