for what value of k the following series will be in arithmetic progression? 3k²+k+1,2k²+k,4k²–6k+1
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Answer:
k = 2 , 1/3
Step-by-step explanation:
The series will be in AP if the difference between the consecutive terms is same ,
2k² + k - (3k² + k + 1) = 4k² - 6k + 1 - (2k² + k)
=> 2k² + k - 3k² - k - 1 = 4k² - 6k + 1 - 2k² - k
=> - k² - 1 = 2k² - 7k + 1
=> 3k² - 7k + 2 = 0
=> (k - 2) (3k - 1) = 0
=> k = 2 , 1/3
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