Question

determine k so that k+2,4k-6 and 3k-2 are 3 consecutive terms of an arithmetic progression

Answer 1

Answer:

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=

4

12

=3

Hence k=3.