# GP - three members

The second and third of a geometric progression are 24 and 12(c+1) respectively, given that the sum of the first three terms of progression is 76 determine value of c

Result

c1 =  2
c2 =  0.333

#### Solution:

$a_{2}=24 \ \\ a_{3}=12(c+1) \ \\ \ \\ q=a_{3}/a_{2}=12(c+1)/24=(c+1)/2 \ \\ \ \\ a_{1}=a_{2}/q=24 \cdot \ 2/(c+1)=48/(c+1) \ \\ \ \\ s=a_{1}+a_{2}+a_{3}=76 \ \\ \ \\ 48/(c+1) + 24 + 12(c+1)=76 \ \\ \ \\ \ \\ 48 + 24(c+1) + 12(c+1)^2=76(c+1) \ \\ 12c^2 -28c +8=0 \ \\ \ \\ p=12; q=-28; r=8 \ \\ D=q^2 - 4pr=28^2 - 4\cdot 12 \cdot 8=400 \ \\ D>0 \ \\ \ \\ c_{1,2}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 28 \pm \sqrt{ 400 } }{ 24 } \ \\ c_{1,2}=\dfrac{ 28 \pm 20 }{ 24 } \ \\ c_{1,2}=1.16666667 \pm 0.83333333333333 \ \\ c_{1}=2 \ \\ c_{2}=0.33333333333333 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 12 (c -2) (c -0.33333333333333)=0 \ \\ \ \\ c=c_{1}=2 \ \\ q=(c+1)/2=(2+1)/2=\dfrac{ 3 }{ 2 }=1.5 \ \\ a_{1}=a_{2}/q=24/1.5=16 \ \\ a_{3}=a_{2} \cdot \ q=24 \cdot \ 1.5=36 \ \\ s_{2}=a_{1}+a_{2}+a_{3}=16+24+36=76 \ \\ \ \\ s_{2}=s \ \\ \ \\ c_{1}=2$

Checkout calculation with our calculator of quadratic equations.

$q=(c_{2}+1)/2=(0.3333+1)/2 \doteq \dfrac{ 2 }{ 3 } \doteq 0.6667 \ \\ a_{11}=a_{2}/q=24/0.6667=36 \ \\ a_{33}=a_{2} \cdot \ q=24 \cdot \ 0.6667=16 \ \\ s_{3}=a_{11}+a_{2}+a_{33}=36+24+16=76 \ \\ s_{3}=s_{2}=s \ \\ c_{2}=0.3333=\dfrac{ 1 }{ 3 } \doteq 0.3333 \doteq 0.333$

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