Question

If the nth term of a progression is (4n – 10) show that it is an AP. Find its (i) first term, (ii) common difference, and (iii) 16th term.

Answer 1

T_n = (4_n – 10)   ...[Given]

T₁ = (4 × 1 – 10) = –6

T₂ = (4 × 2 – 10) = –2

T₃ = (4 × 3 – 10) = 2

T₄ = (4 × 4 – 10) = 6

Clearly, [–2 – (–6)] = [2 – (–2)]

= [6 – 2]

= 4   ...(Constant)

So, the terms –6, –2, 2, 6,...... forms an AP.

Thus we have

(i) First term = –6

(ii) Common difference = 4

(iii) T₁₆ = a + (n – 1)d 

= a + 15d

= –6 + 15 × 4 

= 54