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

1 Answer
Dec 7, 2016

Answer:

#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#