progression It's about a geometric progression, bn, where b1+b2+b3=21 b1*b2*b3=216 It has to be found: b7.
You may also use the property of 3 consecutive terms of geometric progression such that:
b2^2 = b1*b3
The problem provides the information b1*b2*b3=216 => b1*b3 = 216/b2
Substituting 216/b2 for b1*b3 in b2^2 = b1*b3 yields:
b2^2 = 216/b2 => b2^3 = 216 => b2 = 6 => b2^2 = 36 => b1*b3 = 36
But b1+b2+b3=21 => b1+b3 = 21 - 6
b1+b3 = 15
You may form the quadratic equation such that:
x^2 - (b1+b3)x + b1*b3 = 0
x^2 - 15x + 36 = 0
x1 = (15+sqrt(225-144))/2
b3 = (15+9)/2 => b3=12
b1 = (15-9)/2 => b1=3
Hence, you may find the common ratio of progression such that:
q = 12/6 = 6/3 = 2
You may find b7 such that:
b7 = b1*q^6
b7 = 3*2^6 => b7 = 3*64 => b7 = 192
Hence, evaluating b7 yields b7 = 192.
Due to the fact that it's about a geometrical progression;
b2=b1*q, where q is the ratio
In the given relation
b1*b2*b3=216, we'll write b2 and b3 using the relations above, so b1*b2*b3=216 will become
b1*q=2*3, so, from this relation, q could be 2 when b1=3 and reverse.
In the relation b1+b2+b3=21, we could write:
b1+ b1*q + b1*q^2=21
b1*(1+q+q^2)=21, but b1*q=6, so b1=6/q
Given the relation b1*q=2*3, we can conclude that q=q1=2, so b1=3
b1, b2,b3 are in G.P.So,the terms b1, b2 and b3 could be written like: b/r+b and b, where r is the common ratio.
Threfore, b/r+b+br= 21..........(1) and
(b/r)(b)(br)=216............(2), which gives b^3=216.Therefore,
b=216^(1/3) = 6. Substituting this value in (1),
6(1/r+1+r) = 21 or (1/r+r) =3.5 -1 = 2.5 or
r+1/r = 2+1/2. By comparision, r=2.
To find b7:The 1st term is b/r = 6/2 =3.
b7 = (b/r)*r^(7-1) =(6/2)*2^(7-1) = 3*2^6=192