# progression It's about a geometric progression, bn, where b1+b2+b3=21 b1*b2*b3=216 It has to be found: b7.

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### 3 Answers

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;

b1=b1

b2=b1*q, where q is the ratio

b3=b2*q=b1*q*q=b1*q^2

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*(b1*q)*(b1*q^2)=(b1*q)^3=216=(2*3)^3

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

6/q*(1+q+q^2)=21

6/q*1+6/q*q+6/q*q^2=21

6+6q+6q^2=21*q

6q^2-15q+6=0

q1={15+sqrt[(15-12)*(15+12)]}/12=(15+9)/12=24/12=2

q1={15-sqrt[(15-12)*(15+12)]}/12=(15-9)/12=6/12=1/2

Given the relation **b1*q=2*3,** we can conclude that** q=q1=2,** so **b1=3**

**b7=b1*q^6=3*2^6=3*64=192**

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