Question

determine k so that k+2,4k-6 and 3k-2 are three consecutive terms of an A.P.?

Answer 1

according to  arithemetic progression 2b=a+c

then,

2(4k-6)=k+2+3k-2

8k-12=4k

4k=12

k=12/4=3

so,the value of k=3

the thrre terms are k+2=3+2=5

                                4k-6=12-6=6

                                3k-2=9-2=7      

Answer 2

Step-by-step explanation:

(k+2)+(3k−2)=2(4k−6)

⇒4k=8k−12

⇒4k−8k=−12

⇒−4k=−12

⇒4k=12

⇒k=

4

12

=3

Hence k=3.

hope help u