Question

If $a_1, a_2, a_3, \dots a_{2n}$ are in A.P. with common difference d, then $\tan^{-1}\left(\frac{2d}{1+a_1a_3}\right) + \tan^{-1}\left(\frac{2d}{1+a_3a_5}\right) + \tan^{-1}\left(\frac{2d}{1+a_5a_7}\right) + \dots + \tan^{-1}\left(\frac{2d}{1+a_{2n-3}.a_{2n-1}}\right)$ is equal to

• A. $\tan^{-1}(a_{2n-1}) - \tan^{-1}(a_3)$
• B. $\tan^{-1}(a_{2n}) - \tan^{-1}(a_1)$
• C. $\tan^{-1}(a_1) - \tan^{-1}(a_{2n-1})$
• D. $\tan^{-1}(a_{2n-1}) - \tan^{-1}a_1$

Answer 1

Answer: (D)

$\tan^{-1}(a_{2n-1}) - \tan^{-1}(a_1)$

Answer 2

Given an A.P.:

$$a_k = a_1 + (k-1)d$$

For any two odd-indexed terms,

$$a_{2i+1} - a_{2i-1} = [a_1 + 2id] - [a_1 + (2i-2)d] = 2d.$$

Using the formula for the difference of inverse tangents:

$$\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right).$$

Thus,

$$\tan^{-1}(a_{2i+1}) - \tan^{-1}(a_{2i-1}) = \tan^{-1}\left(\frac{2d}{1+a_{2i-1}a_{2i+1}}\right).$$

The sum given is:

$$\sum_{i=1}^{n-1} \tan^{-1}\left(\frac{2d}{1+a_{2i-1}a_{2i+1}}\right).$$

Replacing each term, the sum becomes telescopic:

$$[\tan^{-1}(a_3)-\tan^{-1}(a_1)] + [\tan^{-1}(a_5)-\tan^{-1}(a_3)] + \cdots + [\tan^{-1}(a_{2n-1})-\tan^{-1}(a_{2n-3})].$$

Telescoping gives:

$$\tan^{-1}(a_{2n-1}) - \tan^{-1}(a_1).$$