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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It is given that k+2,4k−6,3k−2 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)±(3k−2)=2(4k−6)
⇒4k=8k−12
⇒4k−8k=−12
⇒−4k=−12
⇒4k=12
⇒k=12/4
=3
Hence k=3.