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

Question

Philoid · Accepted Answer

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