Subharthi's progression

$x^2 - 3x + a = (x - c)(x - d) = x^2 - (c + d)x + cd$ $x^2 - 12x + b = (x - e)(x - f) = x^2 - (e + f)x + ef$

So we have that:

$c + d = 3$

$e + f = 12$

But, like $c, d, e, f$ are consecutive terms of a geometric progression, we know $d = ck$ and $f = dk$ , for some constant $k$ . Rewriting the equations:

$c(1 + k) = 3$

$e(1 + k) = 12$

So,

$e\frac{3} {c} = 12$

$\frac{e} {c} = 4$

But, like $e = ck^2$ , we can write this as:

$\frac{ck^2} {c} = 4$

So, because each term is greater that it's predecessor, we have that k is possitive and $k = 2$ , that means:

$c = 1, b = 2, e = 4, f = 8$

So $a + b = cd + ef = 2 + 32 = 34$

Let cd be equal to a Let ef be equal to b let c be equal to z Therefore, d=zr Therefore, e=zr^2 z^2 \times r=a z^2 \times r^5=b By vieta formula, z+zr=3 zr^2+zr^3=12 Solving these two equations, we get r=positive or negative two If r is positive two, we get z=1 from vietas formula therefore, a+b=30+4=34

Denote $c$ , $d=nc$ , $e=n^2c$ , $f=n^3c$ are the terms of the geometric sequence. Applying Vieta theorem yields: $c+nc=3$ , $nc^2=a$ , $n^2c+n^3c=12$ and $n^5c^2=b$ . From these equation we can find $n=2$ and $c=1$ and therefore, $a+b=34$ .

Let $c = a$ , $d = ar$ , $e = ar^2$ , and $f = ar^3$ . From Vieta's Formulas, we have $c + d = a + ar = 3$ and $e + f = ar^2 + ar^3 = 12$ . Dividing the equations gives $r^2 = 4$ , and since $c<d<e<f$ we have $r=2$ . Substituting gives $a + 2a = 3$ or $a = 1$ , so our geometric sequence is $1, 2, 4, 8$ . Therefore, again from Vieta's, $a = 1 \cdot 2 = 2$ and $b = 4 \cdot 8 = 32$ , so our answer is $32 + 2 = \boxed{34}$ .

Let the common ratio be $r$ . We have, from Vieta's formulas, that $\begin{aligned} c + cr &=& 3, \\ c \cdot cr &=& a, \\ cr^2 + cr^3 &=& 12, \\ cr^2 \cdot cr^3 &=& b. \end{aligned}$ Dividing the first equation from the third, we have $r^2 = 4 \implies r = \pm 2$ .

But the progression is strictly increasing, so $r = 2 \implies c = 1$ . We have $a = c^2 r = 2$ and $b = c^2 r^5 = 32$ , so $a + b = \boxed {34}.$

let c=A, then d=Ar, e=Ar^2, f=Ar^3
3=c+d=A(r+1), 12=e+f=Ar^2(r+1)
solving the equation r=2, A=1
a+b=cd+ef=(A^2)r(1+r^4)=2(17)=34
*actually r=-2 also satisfy the equation but it give u A=-3, a+b =-306