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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»

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



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