Sum of p terms of AP is q and sum of q terms of AP is p. Common difference of A.P. is:

Correct option is B

Given:

Sum of p terms = q

Sum of q terms = p

Formula Used:

​$$S_n = \frac{n}{2}[2a + (n-1)d]$$​

where a is the first term and d is the common difference.

Solution:

From the given,

​$$\frac{p}{2}[2a + (p-1)d] = q$$      (1)

​$$\frac{q}{2}[2a + (q-1)d] = p \quad \text{(2)}$$​

From (1),

​$$2a + (p-1)d = \frac{2q}{p}$$​

From (2),

​$$2a + (q-1)d = \frac{2p}{q}$$​

Subtracting (1) from (2),

​$$[2a + (q-1)d] - [2a + (p-1)d] = \frac{2p}{q} - \frac{2q}{p}$$​

​$$(q-1)d - (p-1)d = \frac{2p}{q} - \frac{2q}{p}$$​

​$$(q - p)d = \frac{2p^2 - 2q^2}{pq}$$​

​$$(q - p)d = \frac{2(p+q)(p-q)}{pq}$$​

​$$d = \frac{-2(p+q)}{pq}$$​​