SOLUTION: a. find the sum of the first n terms of the arithmetic series. b. find n for the given sum S sub n. 45. 3+8+13+18+23+... a. n=20 b. S sub n=366

Question 35878This question is from textbook Algebra 2
: a. find the sum of the first n terms of the arithmetic series.
b. find n for the given sum S sub n.
45. 3+8+13+18+23+...
a. n=20 b. S sub n=366 This question is from textbook Algebra 2

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
a. find the sum of the first n terms of the arithmetic series.
b. find n for the given sum S sub n.
45. 3+8+13+18+23+...
---------------------------------------------
a. n=20
SN=(N/2){2A+(N-1)D}.....A=3....D=8-3=13-8=18-13=23-18=5....N=20
SN=(20/2){2*3+(20-1)5}=10(6+19*5)=10(6+95)=10*101=1010
----------------------------------------------
b. S sub n=366
366=(N/2){2*3+(N-1)5}=(N/2){6+5N-5}=(N/2)(5N+1)
N(5N+1)=366*2=732
5N^2+N-732=0
5N^2+61N-60N-732=0
N(5N+61)-12(5N+61)=0
(5N+61)(N-12)=0
N-12=0...OR...N=12.................5N+61=0 IS NOT TENABLE AS N IS NOT AN INTEGER.
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