Question

What will be the value of k so that (2k + 1), (5k – 3) and (-8k + 5) are three consecutive terms of an AP?

(a) 4

(b) 3

(c) \(\frac {3}{4}\)

(d) \(\frac {4}{3}\)

I got this question by my college director while I was bunking the class.

This interesting question is from Arithmetic Progression Basics in section Arithmetic Progressions of Mathematics – Class 10

Answer 1

Right answer is (c) \(\frac {3}{4}\)

To explain: Since, (2k + 1), (5k – 3) and (-8k + 5) are three consecutive terms of an AP.

The common difference will be same.

(5k – 3) – (2k + 1) = (-8k + 5) – (5k – 3)

3k – 4 =  -13k + 8

16k = 12

k = \(\frac {3}{4}\)