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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)`.
Solution
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
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