```Question 283866
A geometric sequence is a sequence of the form

{{{a}}}, {{{ar}}}, {{{ar^2}}}, {{{ar^3}}}, ....

where 'a' is the starting term and 'r' is the common ratio. If 'a' is positive, then the value of 'r' will determine whether the terms will increase or decrease. If 'r' is greater than 1 (ie {{{r>1}}}), then the terms will increase, while on the other hand, if 'r' is less than 1 ({{{r<1}}}), then the terms will decrease. Finally, if 'r' is equal to 1, then the terms will remain the same.

If {{{r>1}}}, then we can write 'r' as {{{r=1+k}}} where {{{k>1}}}. Plug this into the sequence above to get:

{{{a}}}, {{{a(1+k)}}}, {{{a(1+k)^2}}}, {{{a(1+k)^3}}}, ....

Now expand and distribute

{{{a}}}, {{{a+ak}}}, {{{a+2ak+ak^2}}}, {{{a+3ak^2+3ak+ak^3}}}, ....

Notice how the second term {{{a+ak}}} is bigger than 'a' since 'ak' is positive ('a' is positive and so is 'k'). Symbolically, this can be proven by showing that {{{a<a+ak}}} simplifies to {{{ak>0}}} (which is true, as shown above). Similar arguments can be made to show that {{{a+2ak+ak^2}}} is bigger than {{{a+ak}}}. Similarly, we can show that {{{a+3ak^2+3ak+ak^3}}} is greater than {{{a+2ak+ak^2}}}, and so on.

For {{{r<1}}}, we can write 'r' as {{{r=1-k}}} where {{{0<k<1}}} (we want to keep 'r' positive). I'll let you plug in {{{1-k}}} for 'r'. Use similar arguments shown above to show that each term will get smaller.

Note: if {{{r=1}}}, then the sequence

{{{a}}}, {{{ar}}}, {{{ar^2}}}, {{{ar^3}}}, ....

becomes

{{{a}}}, {{{a(1)}}}, {{{a(1)^2}}}, {{{a(1)^3}}}, ....

which simplifies to

{{{a}}}, {{{a}}}, {{{a}}}, {{{a}}}, ....

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