Let a and b be be two distinct positive real numbers. Let 11th  term of a GP, whose first term is a and third term is b, is equal to pth  term of another GP, whose first term is a and fifth term is b. Then p is equal to

In a geometric progression (GP), the nth term can be found using the formula a*r^(n-1), where a is the first term, r is the common ratio, and n is the term number.

Given that the first term is a and the third term is b in the first GP, we can write the third term as ar^(3-1) = ar^2 = b. From this, we can find the common ratio r = sqrt(b/a).

The 11th term of this GP is ar^(11-1) = ar^10. Substituting r = sqrt(b/a), we get the 11th term as a*(sqrt(b/a))^10 = a*(b/a)^5 = b^5/a^4.

In the second GP, the first term is a and the fifth term is b. So, we can write the fifth term as as^(5-1) = as^4 = b, where s is the common ratio of the second GP. From this, we can find s = (b/a)^(1/4).

The pth term of the second GP is as^(p-1). Substituting s = (b/a)^(1/4), we get the pth term as a((b/a)^(1/4))^(p-1) = a*(b/a)^(p/4-1/4).

Given that the 11th term of the first GP is equal to the pth term of the second GP, we can equate the two expressions:

b^5/a^4 = a*(b/a)^(p/4-1/4)

Solving this equation for p, we get p = 20.