Arithmetic progression problem.

**mastermin346** · July 4th 2011, 02:04 AM

If the $p^{th}$ term of an arithmetic progression is $\frac{1}{q}$ and the $9^{th}$ term $= \frac{1}{p}$ ,show that $(pq)^{th}$ is $1$

**emakarov** · July 4th 2011, 02:51 AM

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.

**mastermin346** · July 4th 2011, 03:08 AM

but my book state that (Pq) th term is 1 sir?

**TKHunny** · July 4th 2011, 03:28 AM

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 $p\cdot q = \frac{1}{(a+8d)(a-d)+d}$

p*q is going to have to be an integer. You tell me how this can be and then the implications of the result.

**Soroban** · July 4th 2011, 05:49 AM

Hello, mastermin346!

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

. . . . . i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\downarrow$
$\text{If the }p^{th}\text{ term of an arithmetic progression is }\tfrac{1}{q}\,\text{ and the }q^{th}\text{ term is }\tfrac{1}{p}$
$\text{show that the }(pq)^{th}\text{ term is }1.$

$\text{Note that: }\,p \ne q$

$\text{The }n^{th}\text{ term is: }\:a_n \:=\:a + (n-1)d$
. . $\text{where }a\text{ = first term, }d\text{ = common difference.}$

$\text{We are told: }\:\begin{Bmatrix} a_p \:=\:\frac{1}{q} & \Rightarrow & a + (p-1)d &=& \frac{1}{q} & [1] \\ \\[-3mm] a_q \:=\:\frac{1}{p} & \Rightarrow & a + (q-1)d &=& \frac{1}{p} & [2] \end{Bmatrix}$

. . $\begin{array}{ccccc}\text{From [1]:} & aq + dpq - dq \:=\:1 & [3] \\ \\[-3mm] \text{From [2]:} & ap + dpq - dp \:=\:1 & [4] \end{array}$

$\text{Equate [3] and [4]: }\:aq + dpq - dq \:=\:ap + dpq - dp$

. . $ap - aq - dp + dq \:=\:0 \quad\Rightarrow\quad a(p-q) - d(p-q) \:=\:0$

. . $(p-q)(a-d)\:=\:0 \quad\Rightarrow\quad \rlap{/////}p = q,\;a = d\;\;[5]$

$\text{Substitute [5] into [3]: }\:dq + dpq - dq \:=\:1 \quad\Rightarrow\quad dpq \:=\:1 \quad\Rightarrow\quad pq \:=\:\tfrac{1}{d}$

$\text{Then: }\:a_{pq} \;=\;a_{\frac{1}{d}} \;=\;d + \left(\tfrac{1}{d}-1\right)d \;=\; d + 1 - d$

. . $\text{Therefore: }\:a_{pq} \;=\;1$

**emakarov** · July 4th 2011, 05:58 AM

Note that: $p\ne q$

This has to be an assumption in the problem statement because nothing prevents $p$ th term to be $1/p$ ; this does not imply that $p^2$ th term is 1.

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