Question

:( I really need help..

Answer 1

option d)
`a(n)=4(n-1)+(3)/(4)`

step by step explanation:

Given sequence is
`a_n={4, (19)/(4),(11)/(2),(25)/(4),7,...}`

Let
`a_1=4`,
`a_2=(19)/(4)`,
`a_3=(11)/(2)`,
`a_4=(25)/(4)`,
`a_5=4`, ...

common difference
`d=a_2-a_1`

`d=(19)/(4)-4`

`d=(19-16)/(4)`

`d=(3)/(4)`

`d=a_3-a_2`

`d=(11)/(2)-(19)/(4)`

`d=(22-19)/(4)`

`d=(3)/(4)`

Therefore the common difference
`d=(3)/(4)`

The recurrsive formula for arithmetic sequence is

`a(n)=a(n-1)+d`

Therefore

`a(n)=4(n-1)+(3)/(4)` where a=4 and
`d=(3)/(4)`