Question

in each of the following geometric sequences, find (i) the 7th term; (ii) the nth term. a. 1/2, 3/4, 9/8, .... b. √5, 5, 5√5, ....c. - 1/2, 3/4, - 9/8, ....

Answer 1

Part a

we have

a. 1/2, 3/4, 9/8, ....

a1=1/2

a2=3/4

a3=9/8

so

a2/a1=(3/4)/(1/2)=3/2

a3/a2=(9/8)/(3/4)=3/2

that means

the common ratio r=3/2

the general formula is

`a_n=a_1\cdot r^((n-1))`

substitute given values

`a_n=(1)/(2)\cdot((3)/(2))^((n-1))`

For n=7

substitute

`\begin{gathered} a_7=(1)/(2)\cdot((3)/(2))^((7-1)) \\ \\ a_7=(1)/(2)\cdot(729)/(64) \\ \\ a_7=(729)/(128) \end{gathered}`

Part b

we have

b. √5, 5, 5√5, ....

a1=√5

a2=5

a3=5√5

so

a2/a1=5/√5=√5

a3/a2=5√5/5=√5

the common ratio is r=√5

the general formula is

`a_n=a_1\cdot r^((n-1))`

substitute given values

`a_n=\sqrt[]{5}_{}\cdot(\sqrt[]{5}^((n-1)))`

For n=7

`\begin{gathered} a_7=\sqrt[]{5}_{}\cdot(\sqrt[]{5}^((7-1))) \\ a_7=125\sqrt[]{5}_{} \end{gathered}`