determine the value of 'k' so that k+2,4k-6,3k-2 are in A.P.
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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=
4
12
=3
Hence k=3.
(k+2)+(3k−2)=2(4k−6)
⇒4k=8k−12
⇒4k−8k=−12
⇒−4k=−12
⇒4k=12
⇒k=
4
12
=3
Hence k=3.
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0
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