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

1.	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» 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

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»

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