The number 11,p,q,21¹/2.....form an A.P. Find p and q

Need answer

To find the values of p and q in the arithmetic progression (A.P.) sequence, let's examine the given terms:

11, p, q, 21¹/2

In an arithmetic progression, the common difference (d) between consecutive terms remains constant. We need to find this common difference to determine the values of p and q.

To find the common difference (d), we need to find the difference between any two consecutive terms in the sequence:

First, let's find the difference between the second term (p) and the first term (11):
p - 11

Next, let's find the difference between the third term (q) and the second term (p):
q - p

Similarly, let's find the difference between the fourth term (21¹/2) and the third term (q):
21¹/2 - q

Since it is an arithmetic progression, the common difference should be the same for all these differences.

So, we can set up the following equations:

p - 11 = q - p
q - p = 21¹/2 - q

Simplifying these equations, we get:

2p = 11 + q
q = 21¹/2 - 2p

Now, we can substitute the value of q from the second equation into the first equation:

2p = 11 + (21¹/2 - 2p)

Expanding and simplifying:

2p = 11 + 21¹/2 - 2p
4p = 11 + 21¹/2

Dividing both sides by 4:

p = (11 + 21¹/2) / 4

Now, we can substitute the value of p back into the equation for q:

q = 21¹/2 - 2p

Substituting the value of p, we can find q.

Answer

q-p = p-11

21.5-q = q-p

p = 29/2, q=18
d = 3 1/2, so the sequence is

11, 14 1/2, 18, 21 1/2