Sequences AP + GP

The three numbers that make up the arithmetic sequence have the sum of 30. If we subtract from the first 5, from the second 4 and keep the third, we get the geometric sequence. Find AP and GP members.

Correct answer:

a1 =  8
b1 =  10
c1 =  12
a2 =  17
b2 =  10
c2 =  3
g1 =  3
g2 =  6
g3 =  12
k1 =  12
k2 =  6
k3 =  3

Step-by-step explanation:

a+b+c=30 b=a+d c=b+d q (a5)=(b4) q (b4)=c  3a+3d=30 a=(303d)/3=10d  b=a+d=10d+d b=10 c=10+d  q (10d5)=6 q 6=10+d  (5d)/6=6/(10+d)=35=0.6  (5d)(10+d)=66  (5d) (10+d)=6 6 d25d+14=0 d2+5d14=0  a=1;b=5;c=14 D=b24ac=5241(14)=81 D>0  d1,2=b±D2a=5±812 d1,2=5±92 d1,2=2.5±4.5 d1=2 d2=7   Factored form of the equation:  (d2)(d+7)=0  a1=bd1=102=8

Our quadratic equation calculator calculates it.

b1=b=10=10
c1=b+d1=10+2=12
a2=bd2=10(7)=17
b2=b=10=10
c2=b+d2=10+(7)=3
g1=a15=85=3
g2=b4=104=6
g3=c1=12=12
k1=a25=175=12
k2=b4=104=6
k3=c2=3   Verifying Solution:  Q1=g2/g1=6/3=2 Q11=g3/g2=12/6=2  Q2=k2/k1=6/12=12=0.5 Q22=k3/k2=3/6=12=0.5



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