How do you find $a_{7}$ $a_7$ for the geometric sequence $1/32, 1/16, 1/8,...$ ?

Question

2396 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-07T10:25:12

$3/8$

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:

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Determine the pattern")$

$a_1" "->" "1/32$

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

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

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

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

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

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

How do you find a_7a7a_7 for the geometric sequence 1/32, 1/16, 1/8,...1/32, 1/16, 1/8,...?