Consecutive 69904

The three numbers that make three consecutive members of an arithmetic sequence have a sum of 60 and a product of 7500. Find these numbers.

Correct answer:

a1 =  25
b1 =  20
c1 =  15

Step-by-step explanation:

a+b+c=60 b=a+d c = a+2d abc = 7500 = 22 × 3 × 54  a+a+d+a+2d = 60 a(a+d)(a+2d) = 7500  3a+3d = 60 a(a+d)(a+2d) = 7500  a+d=20 d = 20a  a(a+20a)(a+2(20a)) = 7500  a 20 (a+2(20a))=7500  a 20 (a+2(20a))=7500 20a2+800a7500=0 20a2800a+7500=0 20=225 800=2552 7500=22354 GCD(20,800,7500)=225=20  a240a+375=0  p=1;q=40;r=375 D=q24pr=40241375=100 D>0  a1,2=2pq±D=240±100 a1,2=240±10 a1,2=20±5 a1=25 a2=15  d1=20a1=2025=5 d2=20a2=2015=5

Our quadratic equation calculator calculates it.

c1=b1+d1=20+(5)=15   Verifying Solution:  s1=a1+b1+c1=25+20+15=60 p1=a1 b1 c1=25 20 15=7500  b2=a2+d2=15+5=20 c2=b2+d2=20+5=25  s2=a2+b2+c2=15+20+25=60 p2=a2 b2 c2=15 20 25=7500

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