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