If the term of an arithmetic progression is and the term ,show that is

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- July 4th 2011, 03:04 AMmastermin346Arithmetic progression problem.
If the term of an arithmetic progression is and the term ,show that is

- July 4th 2011, 03:51 AMemakarovRe: Arithmetic progression problem.
This seems to be false. Take the progression 0, 1/16, 2/16, 3/16, ..., p = 2 and q = 16. Then the second term is 1/16 = 1/q, 9th term is 8/16 = 1/p, but the 32th term is 31/16, not 1.

- July 4th 2011, 04:08 AMmastermin346Re: Arithmetic progression problem.
but my book state that (Pq) th term is 1 sir?

- July 4th 2011, 04:28 AMTKHunnyRe: Arithmetic progression problem.
Books are not written by the great Authority. They are written by humans.

Prove it out. An arithmetic progression has a starting value (call it 'a') and a constant difference (call it 'd').

"the pth term is 1/q"

a + (p-1)d = 1/q

"the 9th term is 1/p"

a + (9-1)d = 1/p

This leads to

p*q is going to have to be an integer. You tell me how this can be and then the implications of the result. - July 4th 2011, 06:49 AMSorobanRe: Arithmetic progression problem.
Hello, mastermin346!

I believe there is a typo . . . It isthe 9th term.*not*

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- July 4th 2011, 06:58 AMemakarovRe: Arithmetic progression problem.Quote:

Note that: