Question

Show that the progression `6, 7 3/4, 9 1/2, 11 1/4`, ......... is an A.P. Write down its (i) first term (ii) common difference (iii) 17th term.

Answer 1

The terms are $$6,\frac{31}{4}, \frac{19}{2}, \frac{45}{4}$$. 

Consecutive differences:

$$\frac{31}{4} - 6 = \frac{7}{4}$$

$$\frac{19}{2} - \frac{31}{4} = \frac{7}{4}$$

$$\frac{45}{4} - \frac{19}{2} = \frac{7}{4}$$

Hence it is an A.P.

Here $$a = 6$$, $$d = \frac{7}{4}$$. 

17th term: $$T_{17} = 6 + (17 - 1) \times \frac{7}{4}$$

$$= 6 + 28$$

$$= 34$$

So first term = 6, common difference = `7/4`, 17th term = 34.