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Determine k so that k²+ 4k + 8, 2k² + 3k + 6, 3k² + 4k + 4 are three consecutive terms of an AP

Solution:

An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same.

It is given that

k² + 4k + 8, 2k² + 3k + 6 and 3k² + 4k+ 4 are consecutive terms of an AP.

2k²+ 3k + 6- (k² + 4k + 8)

= 3k² + 4k + 4 - (2k² + 3k + 6) is the Common difference

By multiplying the negative sign

2k² + 3k + 6 - k² - 4k - 8 = 3k² + 4k + 4 - 2k² - 3k - 6

k² - k - 2 = k² + k - 2

-k = k

2k = 0

k = 0

Therefore, k is 0.

✦ Try This: Determine k so that k²+ 2k + 6, 4k² + 5k + 6, 2k² + 5k + 5 are three consecutive terms of an AP

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5

NCERT Exemplar Class 10 Maths Exercise 5.3 Problem 11

Summary:

k is 0 so that k²+ 4k + 8, 2k² + 3k + 6, 3k² + 4k + 4 are three consecutive terms of an AP.

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