Roots in a geometric progression

Let the common ratio be $r$ . Then $b = ar$ and $c = ar^{2}$ . By Vieta's we have that

$a + b = 12, ab = p$ and $b + c = 15, bc = q$ .

Thus $a + ar = a(1 + r) = 12$ and $ar + ar^{2} = ar(1 + r) = 15$ , and so

$\dfrac{ar(1 + r)}{a(1 + r)} = \dfrac{15}{12} \Longrightarrow r = \dfrac{5}{4}$ .

We then have that $a = \dfrac{12}{1 + r} = \dfrac{12}{1 + \dfrac{5}{4}} = \dfrac{48}{9} = \dfrac{16}{3}$ .

Then $b = \dfrac{16}{3}*\dfrac{5}{4} = \dfrac{20}{3}$ and $c = \dfrac{16}{3}*\dfrac{25}{16} = \dfrac{25}{3}$ .

Finally, $p + q = ab + bc = b(a + c) = \dfrac{20}{3}*\left(\dfrac{16}{3} + \dfrac{25}{3}\right) = \dfrac{20*41}{9} = \boxed{91.11}$ to 2 decimal places.

we know that a+b=12 and b+c=15, so we know a+2b+c=27

I thought the number with the ratio 2, with a=3, then b=6, c=12

so we could prove that a+2b+c=27 (3)+2(6)+12=3+12+12=27

so know we move to p and q. we learn that p=ab and q=cb,

thus p+q=ab+bc=b(a+c)=6(3+12)=6(15)=90

Did I do something wrong? cause my answer was also correct :3

a+b=12. eq1 ab=p

b+c=15. eq2 bc=q

a,b,c are in GP Therefore, b²=ac Eq3

Adding eq1 and eq2 gives

a+c+2b=27 Eq4

By Eq3 c=b²/a (Substite c in Eq 4)

a+(b²/a)+2b=27(multiply 'a' on both side)

a²+b²+2ab=27a

Therefore, (a+b)²=27a (but a+b =12)

Therefore a=144/27

→a=16/3 , b=20/3 , c=25/3

p+q=b(a+c)

→p+q=820/9

→p+q=91.1111111