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Rajan@2021
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Determine k so that k+2, 4k-6 and 3k-2 are three consecutive terms of A.P.

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We have a question from arithmetic progression chapter in which we are to find the value of k so that k+2, 4k-6 and 3k-2 are three consecutive terms of A.P.

Book – RS Aggarwal, class 10, chapter 5B,  question no 1

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  1. It is given that k+2,4k6,3k2 are the consecutive terms of A.P, therefore, by arithmetic mean property we have, first term+third term is equal to twice of second term that is:(k+2)±(3k2)=2(4k6)

    4k=8k12

    4k8k=12

    4k=12

    4k=12

    k=12/4

          ​=3

    Hence k=3.

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