Question
Let $a_{1},a_{2},a_{3},\ldots\ldots,a_{n}$ be an arithmetic progression with common difference 'd'. Then $\sum_{k = 1}^{n - 1}\mspace{2mu}\frac{1}{a_{k}a_{k + 1}}$ is equal to
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Solution
Sol. [Key Concept: For an A.P., $\frac{1}{a_{k}\ a_{k + 1}} = \frac{1}{d}(\frac{1}{a_{k}} - \frac{1}{a_{k + 1}})$ telescopes to $\frac{n}{a_{1\ }a_{n + 1}}$.]
$$\sum_{k = 1}^{n - 1}\mspace{2mu}\frac{1}{a_{k}a_{k + 1}} = \frac{1}{\text{ }d}\sum_{k = 1}^{n - 1}\mspace{2mu}\frac{a_{k + 1} - a_{k}}{a_{k}a_{k + 1}} = \frac{1}{\text{ }d}\sum_{k = 1}^{n - 1}\mspace{2mu}\left( \frac{1}{a_{k}} - \frac{1}{a_{k + 1}} \right)$$
$$= \frac{1}{\text{ }d}\left( \frac{1}{a_{1}} - \frac{1}{a_{n}} \right) = \frac{(n - 1)}{a_{1}a_{n}} = \frac{(n - 1)}{a_{1}\left( a_{1} + (n - 1)d \right)}$$
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