Find the common difference and 99^th term of the arithmetic progression :

$7\dfrac{3}{4}, 9\dfrac{1}{2}, 11\dfrac{1}{4}, …….$

The sequence = $\dfrac{31}{4}, \dfrac{19}{2}, \dfrac{45}{4}$ …….

Common difference = $\dfrac{19}{2} - \dfrac{31}{4}$

$\phantom{Common difference}= \dfrac{38 - 31}{4} \\[1em] \phantom{Common difference}= \dfrac{7}{4} \\[1em] \phantom{Common difference}= 1\dfrac{3}{4}$

We know that n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

So, 99^th term,

$a_{99} = \dfrac{31}{4} + (99 - 1) \times \dfrac{7}{4} \\[1em] = \dfrac{31}{4} + \dfrac{98 \times 7}{4} \\[1em] = \dfrac{31}{4} + \dfrac{686}{4} \\[1em] = \dfrac{717}{4} \\[1em] = 179\dfrac{1}{4}.$

Hence, 99^th term of the sequence = $179\dfrac{1}{4}$ and common difference = $1\dfrac{3}{4}$.