Show that the progression 6, $7\dfrac{3}{4}, 9\dfrac{1}{2}, 11\dfrac{1}{4}$,……..is an A.P. Write down its

(i) first term

(ii) common difference

(iii) 17th term.

$7\dfrac{3}{4} = \dfrac{31}{4} = 7.75 \\[1em] 9\dfrac{1}{2} = \dfrac{19}{2} = 9.5 \\[1em] 11\dfrac{1}{4} = \dfrac{45}{4} = 11.25 \\[1em]$

Since, 7.75 - 6 = 1.75, 9.5 - 7.75= 1.75 and 11.25 - 9.5 = 1.75.

Hence, the series is an A.P. with common difference = 1.75.

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.

⇒ a₁₇ = 6 + (17 - 1)1.75

= 6 + (16)1.75

= 6 + 28

= 34.

Hence, a = 6, d = $\dfrac{7}{4}$ and a₁₇ = 34.