In the infinite sequence

a, b, (a + b)/2, (a + 3b)/4, . . .

each term after the second is the arithmetic mean of the two previous terms.

Find the limit of the sequence in terms of real numbers a and b.

Let us denote the nth term of the give sequence as S(n).

Then, we observe that:

S(1) = a

S(2) = b

S(3) = a/2 + b/2

S(4) = a/4 + 3*(b/4)

S(5) = 3*(a/8) + 5*(b/8)

S(6) = 5*(a/16) + 11*(b/16)

S(7) = 11*(a /32) + 21*(b/32)

S(8) = 21*(a/64) + 43*(b/64)

S(9) = 43*(a/128) + 85*(b/128), and so on.

It can be easily shown by means of analytical procedure that;

S(2j) = [1- (1/4)^(j-1)]*(a/3) +[2 + (1/4)^(j-1)]*(b/3); and:

S(2j - 1) = [1+ (1/2)^(2j-3)]*(a/3) +[2 - (1/2)^(2j-3)]*(b/3)

whenever j = 1,2,3, 4, ....

In other words,

S(p)

= [1- (1/4)^(p/2 -1)]*(a/3) +[2 + (1/4)^(p/2-1)]*(b/3);

whenever p is even

= [1+ (1/2)^(p-2)]*(a/3) +[2 - (1/2)^(p-2)]*(b/3);

whenever p is odd

Accordingly,

Limit (p-> infinity) S(p) = a/3 + 2*(b/3)

Consequently, the required limit is (a+2b)/3