Explanation

(c)
Since, $a_{1},\ a_{2},\ a_{3},\ldots,a_{n}$ are in AP with common difference 5
$\Rightarrow \ \ a_{2} - a_{1} = a_{3} - a_{2} = \ldots = a_{n} - a_{n - 1} = 5$
Now $T_{1} = \tan^{- 1}\frac{5}{1 + a_{1}a_{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\frac{a_{2} - a_{1}}{1 + a_{2}a_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}{a_{2} - \tan^{- 1}a_{1}}$
Similarly
$\ \ \ \ \ \ \ \ \ \ \ \ T_{2} = \tan^{- 1}{a_{3} - \tan^{- 1}a_{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ T_{3} = \tan^{- 1}{a_{4} - \tan^{- 1}a_{3}}$
$\ \ \ \ \ \ \ T_{n - 1} = \tan^{- 1}{a_{n} - \tan^{- 1}a_{n - 1}}$
On adding all, we get
$\therefore$ Required sum$= \tan^{- 1}{a_{n} - \tan^{- 1}a_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\frac{a_{n} - a_{1}}{1 + a_{n}a_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\frac{a_{1} + 5(n - 1) - a_{1}}{1 + a_{n}a_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\frac{5(n - 1)}{1 + a_{n}a_{1}}$