Question

How do you find `a_7` for the geometric sequence `1/32, 1/16, 1/8,...`?

Answer 1

`3/8`

Explanation:

Let the ith term be `a_i`

Note that:`" " 1/32<1/16<1/8`

`color(blue)("First thoughts")`

So the first term is: `" "a_i->a_1=1/32`

Notice that `16=32/2" and that "8=16/2`

To change `1/32` into `1/16` we multiply by 2 in that:

`1/32xx2 = 2/32 = (2-:2)/(32-:2) =1/16`

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`color(blue)("Determine the pattern")`

`a_1" "->" "1/32`

`a_2" "->" "1/32xx2 = 1/16`

`a_3" "->" "1/32xx2xx2=1/8`

By observation we spot that for any `i` we have:

`a_i=1/32xx2(i-1)`

Thus `" "a_i->a_7=1/32xx2(7-1)" "=" "1/32xx12" "=" "12/32`

`12/32" "-=" "(12-:4)/(32-:4) " "=" " 3/8`