If $a_1, a_2, a_3, \ldots, a_n$ is an arithmetic progression with common difference $d$, then evaluate the following expression: $ an\left[ an^{-1}\frac{d}{1+a_1a_2}+ an^{-1}\frac{d}{1+a_2a_3}+ an^{-1}\frac{d}{1+a_3a_4}+\cdots+ an^{-1}\frac{d}{1+a_{n-1}a_n} ight]$.

Question

If $a_1, a_2, a_3, \ldots, a_n$ is an arithmetic progression with common difference $d$, then evaluate the following expression: $	an\left[	an^{-1}\frac{d}{1+a_1a_2}+	an^{-1}\frac{d}{1+a_2a_3}+	an^{-1}\frac{d}{1+a_3a_4}+\cdots+	an^{-1}\frac{d}{1+a_{n-1}a_n}ight]$.

Accepted Answer

> [!TLDR]  
> The key idea is that each term $	an^{-1}\frac{d}{1+a_k a_{k+1}}$ telescopes into $	an^{-1} a_{k+1} - 	an^{-1} a_k$ using the formula for $	an^{-1} x - 	an^{-1} y$. The sum collapses to $	an^{-1} a_n - 	an^{-1} a_1$, and the final tangent simplifies to $\frac{(n-1)d}{1 + a_1 a_n}$.

We have an arithmetic progression $a_1, a_2, \ldots, a_n$ with common difference $d$. So $a_{k+1} = a_k + d$ for each $k$.

The expression inside the outer $	an$ is a sum of arctangents. The trick is to rewrite each term so that consecutive terms cancel.

Recall the identity for the difference of two arctangents:
$$
	an^{-1} x - 	an^{-1} y = 	an^{-1} \frac{x - y}{1 + xy}
$$
provided $xy > -1$ (which holds here for typical AP values, but we proceed formally).

Notice that for any two consecutive terms $a_k$ and $a_{k+1}$, we have $a_{k+1} - a_k = d$. So
$$
	an^{-1} a_{k+1} - 	an^{-1} a_k = 	an^{-1} \frac{d}{1 + a_k a_{k+1}}.
$$

That is exactly the $k$-th term of the sum! So each term in the sum is a difference:
$$
	an^{-1} \frac{d}{1 + a_k a_{k+1}} = 	an^{-1} a_{k+1} - 	an^{-1} a_k.
$$

Now the whole sum becomes:
$$
\sum_{k=1}^{n-1} \left( 	an^{-1} a_{k+1} - 	an^{-1} a_k ight).
$$

This is a telescoping series. Write it out:
- For $k=1$: $	an^{-1} a_2 - 	an^{-1} a_1$
- For $k=2$: $	an^{-1} a_3 - 	an^{-1} a_2$
- ...
- For $k=n-1$: $	an^{-1} a_n - 	an^{-1} a_{n-1}$ …

Q.If $a_1, a_2, a_3, \ldots, a_n$ is an arithmetic progression with common difference $d$ , then evaluate the following expression: $\tan\left[\tan^{-1}\frac{d}{1+a_1a_2}+\tan^{-1}\frac{d}{1+a_2a_3}+\tan^{-1}\frac{d}{1+a_3a_4}+\cdots+\tan^{-1}\frac{d}{1+a_{n-1}a_n}\right]$ .

Concept understanding — Inverse Tangent Identity

Inverse Tangent Identities

The core addition identity

Subtraction

The doubling identity

The complementary identity

Method: Telescoping a sum of arctangents

Steps

Common Mistakes

The core idea: the range of $\tan^{-1}$ and the product condition

Applying it to this problem

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