You can put this solution on YOUR website! Find the first three terms of the arithmetic series in which asub1= 5 asubn= 33 and Ssubn= 152
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a(1)=5
a(n)=a(1)+(n-1)d=33
5+(n-1)d=33
(n-1)d=28
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S(n)=(n/2)(a(1)+a(n))=152
(n/2)(5+33)=152
n*38=304
n=8
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Since n=8 and (n-1)d=28
7d=28
d=4
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So,
a(1)=5
a(2)=5+d=5+4=9
a(3)5+2d=5+8=13
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Cheers,
Stan H.