Question

Show that the progression 10, 6, 2, –2, –6, ......... is an A.P. Write down its (i) first term (ii) common difference (iii) 10th term.

Answer 1

Consecutive differences are equal:

$$6 - 10 = 2 - 6 = -2 - 2 = -6 - (-2) = -4$$

Hence it is an A.P.

Here $$a = 10$$, $$d = -4$$. 

10th term: $$T_{10} = 10 + (10 - 1) \times (-4)$$

$$= 10 - 36$$

$$= -26$$

So first term = 10, common difference = –4, 10th term = –26.