Six speeds

A drilling machine is to have 6 speed ranging from 50 to 750 revolution per minute. If the speed forms a geometric progression, determine their values.

Result

a1 =  50 rpm
a2 =  85.939 rpm
a3 =  147.71 rpm
a4 =  253.88 rpm
a5 =  436.362 rpm
a6 =  750.007 rpm

Solution:

a1=50 rpma_{1}=50 \ \text{rpm}
a6=750 rpm  a6=q5 a1  q=a6/a15=750/505≐1.7188   a2=q⋅ a1=1.7188⋅ 50≐85.9386≐85.939 rpma_{6}=750 \ \text{rpm} \ \\ \ \\ a_{6}=q^5 \ a_{1} \ \\ \ \\ q=\sqrt[5]{ a_{6}/a_{1}}=\sqrt[5]{ 750/50} \doteq 1.7188 \ \\ \ \\ \ \\ a_{2}=q \cdot \ a_{1}=1.7188 \cdot \ 50 \doteq 85.9386 \doteq 85.939 \ \text{rpm}
a3=q⋅ a2=1.7188⋅ 85.9386≐147.7095≐147.71 rpma_{3}=q \cdot \ a_{2}=1.7188 \cdot \ 85.9386 \doteq 147.7095 \doteq 147.71 \ \text{rpm}
a4=q⋅ a3=1.7188⋅ 147.7095≐253.8798≐253.88 rpma_{4}=q \cdot \ a_{3}=1.7188 \cdot \ 147.7095 \doteq 253.8798 \doteq 253.88 \ \text{rpm}
a5=q⋅ a4=1.7188⋅ 253.8798≐436.3618≐436.362 rpma_{5}=q \cdot \ a_{4}=1.7188 \cdot \ 253.8798 \doteq 436.3618 \doteq 436.362 \ \text{rpm}
a6=q⋅ a5=1.7188⋅ 436.3618≐750.0068≐750.007 rpma_{6}=q \cdot \ a_{5}=1.7188 \cdot \ 436.3618 \doteq 750.0068 \doteq 750.007 \ \text{rpm}

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