Question

If the pth term of an A. P. is `1/q` and qth term is `1/p`, prove that the sum of first pq terms of the A. P. is `((pq+1)/2)`.

Answer 1

Suppose a be the first term and d be the common difference of the given AP

`a_p = 1/q`

`=> a + (p -1)d = 1/q`  ....(1)

And

`a_q = 1/p`

`=> a + (q - 1)d  = 1/p` ....(2)

Subtracting (2) from (1), we get

`1/q -  1/p= (p - q)d`

`=> (p - q)/"pq" = (p - q)d`

`=>d = 1/(pq)`

Putting d = `1/"pq"` in 1 we get

`a + (p - 1) 1/"pq" = 1/q`

`=> a + 1/q - 1/"pq" = 1/q`

`=> a = 1/"pq"`

∴ Sum of pq terms,

`S_"pq" = "pq"/2 [2a + (pq - 1)d]`

`= "pq"/2[2/"pq" + (pq - 1) 1/"pq"]`

`= "pq"/2 ((1 + pq)/(pq))`

`=((pq + 1)/2)`

Hence proved