In a A.P. and an H.P. have the same first term, the same last term and the same number of terms: Prove that the product of the $r^{t h}$ term from the beginning in one series and the $r^{t h}$ term from the end in the other is independent of $r$ .

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Solution

For the $A . P .$ , $1 s t$ term $= p,$ $n t h$ term $= q$
$q = p + (n - 1) d,$ where $d$ is the common difference
$\Rightarrow d = \frac{q - p}{(n - 1)}$
$(r - 1) t h$ term of the $A . P .$ $= p + \frac{r (q - p)}{(n - 1)} = \frac{p (n - 1) + r (q - p)}{(n - 1)}$ ...(1)
For $H . P .$ $1 s t$ term $= p,$ $n t h$ term $= q$
$\Rightarrow$ the corresponding $A . P .$ has the first term $\frac{1}{p}$ and the $n t h$ term $\frac{1}{q}$
$\frac{1}{q} = \frac{1}{p} + (n + 1) D,$ whre $D$ is the common difference
$\Rightarrow D = \frac{\frac{1}{q} - \frac{1}{p}}{(n - 1)} = \frac{(p - q)}{p q (n - 1)}$
$(n - r) t h$ term of the corresponding $A . P .$
$= \frac{1}{p} + \frac{(n - r - 1) (p - q)}{p q (n - 1)} = \frac{q (n - 1) + (n - r - 1) (p - q)}{p q (n - 1)}$
$= \frac{(n - r - 1) p + q r}{p q (n - 1)} = \frac{p (n - 1) + r (q - p)}{p q (n - 1)}$
Therefore. The $(n - r t h$ term of the $H . P .$ $= \frac{p q (n - 1)}{p (n - 1) + r (q - p)}$ ...(2)
From (1) and (2) we see that the product of the terms $= p q$ independent of $r$

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In a A.P. and an H.P. have the same first term, the same last term and the same number of terms: Prove that the product of the rth term from the beginning in one series and the rth term from the end in the other is independent of r.

In a A.P. and an H.P. have the same first term, the same last term and the same number of terms: Prove that the product of the $r^{t h}$ term from the beginning in one series and the $r^{t h}$ term from the end in the other is independent of $r$ .