Q69 / 150 · MHT CET 2026 (20 April Shift 1) · Mathematics
If $a_1, a_2, a_3, \ldots, a_n$ are in arithmetic progression with common difference d, then $\tan\left[\tan^{-1}\left(\dfrac{d}{1 + a_1 a_2}\right) + \tan^{-1}\left(\dfrac{d}{1 + a_2 a_3}\right) + \ldots + \tan^{-1}\left(\dfrac{d}{1 + a_{n-1} a_n}\right)\right] = $ ____
- A$\dfrac{a_1 - a_n}{1 + a_1 a_n}$
- B$\dfrac{a_n - a_1}{1 - a_1 a_n}$
- Correct answerC$\dfrac{a_n - a_1}{1 + a_1 a_n}$
- D$\dfrac{a_1 + a_n}{1 + a_1 a_n}$
Solution
Given $a_1, a_2, \ldots, a_n$ are in arithmetic progression with common difference $d$.
$d = a_2 - a_1 = a_3 - a_2 = \ldots = a_n - a_{n-1}$
The general term of the series is:
$\tan^{-1}\left(\dfrac{d}{1 + a_k a_{k+1}}\right) = \tan^{-1}\left(\dfrac{a_{k+1} - a_k}{1 + a_k a_{k+1}}\right)$
Using the identity $\tan^{-1}\left(\dfrac{x - y}{1 + xy}\right) = \tan^{-1} x - \tan^{-1} y$, we get:
$\tan^{-1}\left(\dfrac{a_{k+1} - a_k}{1 + a_k a_{k+1}}\right) = \tan^{-1} a_{k+1} - \tan^{-1} a_k$
Let $S$ be the sum of the inverse trigonometric series:
$S = \sum_{k=1}^{n-1} \tan^{-1}\left(\dfrac{d}{1 + a_k a_{k+1}}\right)$
$S = (\tan^{-1} a_2 - \tan^{-1} a_1) + (\tan^{-1} a_3 - \tan^{-1} a_2) + \ldots + (\tan^{-1} a_n - \tan^{-1} a_{n-1})$
This is a telescoping sum, which simplifies to:
$S = \tan^{-1} a_n - \tan^{-1} a_1$
Applying the inverse tangent difference formula:
$S = \tan^{-1}\left(\dfrac{a_n - a_1}{1 + a_1 a_n}\right)$
The given expression is $\tan(S)$:
$\tan(S) = \tan\left[\tan^{-1}\left(\dfrac{a_n - a_1}{1 + a_1 a_n}\right)\right] = \dfrac{a_n - a_1}{1 + a_1 a_n}$
$d = a_2 - a_1 = a_3 - a_2 = \ldots = a_n - a_{n-1}$
The general term of the series is:
$\tan^{-1}\left(\dfrac{d}{1 + a_k a_{k+1}}\right) = \tan^{-1}\left(\dfrac{a_{k+1} - a_k}{1 + a_k a_{k+1}}\right)$
Using the identity $\tan^{-1}\left(\dfrac{x - y}{1 + xy}\right) = \tan^{-1} x - \tan^{-1} y$, we get:
$\tan^{-1}\left(\dfrac{a_{k+1} - a_k}{1 + a_k a_{k+1}}\right) = \tan^{-1} a_{k+1} - \tan^{-1} a_k$
Let $S$ be the sum of the inverse trigonometric series:
$S = \sum_{k=1}^{n-1} \tan^{-1}\left(\dfrac{d}{1 + a_k a_{k+1}}\right)$
$S = (\tan^{-1} a_2 - \tan^{-1} a_1) + (\tan^{-1} a_3 - \tan^{-1} a_2) + \ldots + (\tan^{-1} a_n - \tan^{-1} a_{n-1})$
This is a telescoping sum, which simplifies to:
$S = \tan^{-1} a_n - \tan^{-1} a_1$
Applying the inverse tangent difference formula:
$S = \tan^{-1}\left(\dfrac{a_n - a_1}{1 + a_1 a_n}\right)$
The given expression is $\tan(S)$:
$\tan(S) = \tan\left[\tan^{-1}\left(\dfrac{a_n - a_1}{1 + a_1 a_n}\right)\right] = \dfrac{a_n - a_1}{1 + a_1 a_n}$
Hence, the correct answer is option: 'C'.