SOLUTION: how many terms of the series 24,20,16...must be taken that the sum may be 72?

Question 762889: how many terms of the series 24,20,16...must be taken that the sum may be 72?
Answer by sachi(548) (Show Source): You can put this solution on YOUR website!
the series is 24,20,16...in AP
so a=1st term=24 & common difference =20-24=-4
sum=72 is of say n terms
then 72=n/2[2*24+(n-1)*-4]=n/2[48-4n+4]=n/2[52-4n]
or 144=52n-4n^2
or 4n^2-52n+144=0
or n^2-13n+36=0
or n^2-9n-4n+36=0
or n(n-9)-4(n-9)=0
or (n-9)(n-4)=0
or n=9 or 4

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