Two geometric progressions

Insert several numbers between numbers 6 and 384 so that they form with the given GP numbers and that the following applies:
a) the sum of all numbers is 510

And for another GP to apply:

b) the sum of entered numbers is -132
(These are two different geometric sequences, but with the same two members)

Correct answer:

n =  4
q =  4
n2 =  7
q2 =  -2

Step-by-step explanation:

a1=6 an=384 sn=510  sn=a1 qn1q1 an=a1 qn/q  sn=a1 an/a1 q1q1  sn (q1)=a1 (an/a1 q1) 510 (q1)=6 (384/6 q1)  756q=3024  q=4  n=log(an/a1 q)/log(q)=log(384/6 4)/log(4)=4
s2=132+a1+an=132+6+384=258  s2 (q21)=a1 (an/a1 q21) 258 (q21)=6 (384/6 q21)  756q2=1512  q2=2  n2=7   Verifying Solution:  b1=a1=6 b2=b1 q2=6 (2)=12 b3=b2 q2=(12) (2)=24 b4=b3 q2=24 (2)=48 b5=b4 q2=(48) (2)=96 b6=b5 q2=96 (2)=192 b7=b6 q2=(192) (2)=384 b7=an  S2=b2+b3+b4+b5+b6=(12)+24+(48)+96+(192)=132  n2=log(an/a1 q2)/log(q2)=7



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