sequence is defined by un+1 =p*un +q, where p and q are constants.

[petr]

hello,

The nth term of a sequence is un .
The sequence is defined by u_n+1 =p*u_n +q, where p and q are constants.
The second term of the sequence is 160.
The third term of the sequence is 132 .
The limit of un as n tends to infinity is 20 .

(a) Find the value of p and the value of q
132 =160p +q
20 = 20p+q <------ how is this equation created? why can they use the sum to infinity?

u_n tends to infinity so the last value would be 20 therefore, I can say 20 = p*u_n+q and the Un in this case is twenty again because the penultimate value is close to 20? right?

thank you

 

[pka]

The nth term of a sequence is un .
The sequence is defined by u_n+1 =p*u_n +q, where p and q are constants.
The second term of the sequence is 160.
The third term of the sequence is 132 .
The limit of un as n tends to infinity is 20 .
(a) Find the value of p and the value of q
132 =160p +q
20 = 20p+q <------ how is this equation created? why can they use the sum to infinity?

If $\displaystyle \large u_n\to 20$ THEN SO DOES $\displaystyle \large u_{n+1}\to 20$ thus you get $\displaystyle \large 20=20p+q$.

At this point, you show of your own work.

 

[stapel]

The sequence is defined by u_n+1 =p*u_n +q, where p and q are constants.
The second term of the sequence is 160.
The third term of the sequence is 132 .
The limit of un as n tends to infinity is 20 .

(a) Find the value of p and the value of q
132 =160p +q
20 = 20p+q <------ how is this equation created?

They gave you the formula for the general (that is, the n-th) term. They gave you the values for the 2nd (that is, the n=2) term and the 3rd (that is, the n= 3) term. Plug the given values into the given formula, like you learned back in algebra.