Question

f) If k, k^2
+ 1 and k + 6 are in A.P., find k.​

Answer 1

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=