Question

Find the nth term of the geometric sequence where a_1=4 and a_k+1=(1/2)a_k

Answer 1

Hm, let's see if we can find a pattern (I dunno where I'm going with this, but we'll see!):
\[a_{1}=4\]\[a_{k+1} = {1\over2}a_{k}\]It looks like therefore, \[a_{2} = {1 \over 2} a_{1} = {1\over2} \times 4 = 2\]Likewise, \[a_{3} = {1\over2} \times 2 = 1\]If you mull over this a bit, you can see that each term gets smaller by a factor of one-half. And since the initial value is four, we can guess:
\[a_{n}= 4({1 \over 2})^{n}\]And try substituting in n = 1,2,3. You can see that it actually DOESN'T WORK, everything is too small by a factor of one-half! So let's guess:\[a_{n} = 4 ({1\over2})^{n-1}\]Now try it--this equation should work!

Answer 2

a_k+1=(1/2)a_k==>q=a_k+1/a_k=1/2
a_n=a_1*q^n-1==>a_n=4*(1/2)^n-1

Answer 3

I appreciate your help :)
thank you!

Answer 4

can u give a BA