Q11 of 84 Page 5

If k, 2k –1 and 2k + 1 are three consecutive terms of an AP, the value of k is (CBSE 2014)

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 first three terms of an A.P. respectively are 3y – 1, 3y + 5 and 5y + 1. Then, y equals (CBSE 2014)

The next term of the A.P. (CBSE 2014)

For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A. P.? (CBSE 2016)

Find the 9^th term from the end (towards the first term) of the A.P. 5, 9, 13, …, 185. (CBSE 2016)