Q40 of 202 Page 10

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

The common difference of the A.P. is

The next term of the A.P.

The first three terms of an A.P. respectively are 3y – 1, 3y + 5 and 5y + 1. Then, y equals