Quadrenergic

Of the positive numbers 32, a, b, 128, the first three are three consecutive terms of an arithmetic sequence, the last three are three consecutive terms of a geometric sequence. Determine the value of the terms a and b.

Final Answer:

a =  59.7128
b =  87.4256

Step-by-step explanation:

$$\begin{gathered} a=32+d \\ b=32+2d \\ b=qa \\ 128=q^{2}a \\ b=32+a \\ {b}^2=128a \\ d=a-32=b-a \\ a-32=b-a \\ 2a=b+32 \\ b=2a-32 \\ {b}^2=128a \\ (2a-32{)}^2=128a \\ (2a-32{)}^2=128a \\ 4a^2-256a+1024=0 \\ 4=2^2 \\ 256=2^8 \\ 1024=2^{10} \\ \text{GCD}(4,256,1024)=2^2=4 \\ {a}^2-64a+256=0 \\ p=1;q=-64;r=256 \\ D=q^2-4pr=64^2-4\cdot 1\cdot 256=3072 \\ D>0 \\ {a}_{1,2}=\frac{-q\pm \sqrt{D}}{2p}=\frac{64\pm \sqrt{3072}}{2}=\frac{64\pm 32\sqrt{3}}{2} \\ {a}_{1,2}=32\pm 27.712813 \\ {a}_1=59.712812921 \\ {a}_2=4.287187079 \\ a=a_1=59.7128 \end{gathered}$$

Our quadratic equation calculator calculates it.

$$\begin{gathered} b=2\cdot a-32=2\cdot 59.7128-32\doteq 87.4256 \\ \text{ Verifying Solution: } \\ {d}_1=a-32=59.7128-32=16\sqrt{3}\doteq 27.7128 \\ {d}_2=b-a=87.4256-59.7128=16\sqrt{3}\doteq 27.7128 \\ {q}_1=b/a=87.4256/59.7128\doteq 1.4641 \\ {a}_2=128/b=128/87.4256\doteq 1.4641 \end{gathered}$$

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