Question

If k, 2k –1 and 2k + 1 are three consecutive terms of an AP, the value of k is 

A. – 2 

B. 3 

C. – 3 

D. 6

Answer 1

Here A.P = k, 2k –1, 2k + 1 

Since the numbers are in A.P their common difference (d) should be same 

d=a₂–a₁ = a₃–a₂ 

2k–1–k = 2k + 1– (2k –1) 

k– 1 = 2 

k = 3