If a_1,a_2,,a_n are in H.P., then the expression a_1a_2+a_2a_3++a_n-1a_n is equal to:

Question

If $a_1,a_2,\ldots,a_n$ are in H.P., then the expression $a_1a_2+a_2a_3+\cdots+a_{n-1}a_n$ is equal to:

Accepted Answer

Since $a_1,\ldots,a_n$ are in H.P., $\dfrac{1}{a_1},\ldots,\dfrac{1}{a_n}$ are in A.P. with common difference $d$, so $\dfrac{1}{a_{k+1}}-\dfrac{1}{a_k}=d$, i.e. $a_ka_{k+1}=\dfrac{a_k-a_{k+1}}{d}$.

Summing for $k=1$ to $n-1$ (telescoping):
$$\sum_{k=1}^{n-1}a_ka_{k+1}=\frac{a_1-a_n}{d}.$$

Also $d=\dfrac{1}{a_n}-\dfrac{1}{a_1}\cdot\dfrac{1}{n-1}$... more directly, $d=\dfrac{1}{n-1}\left(\dfrac{1}{a_n}-\dfrac{1}{a_1}\right)=\dfrac{a_1-a_n}{(n-1)a_1a_n}$, so
$$\frac{a_1-a_n}{d}=\frac{a_1-a_n}{\frac{a_1-a_n}{(n-1)a_1a_n}}=(n-1)a_1a_n.$$

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