Question

Determine k so that ( 3k-2), (4k-6) and (k+2) are three consecutive terms of an AP.

Answer 1

Given:( 3k-2), (4k-6) and (k+2) are three consecutive terms of an AP.

To find The value of k

Let a= 3k-2

a2= 4k -6

a+d = 4k -6

3k-2 +d = 4k -6

d = k-4

a3 = k+2

a+ 2d = k+2

3k-2 +2k -8= k+2

5k -10 = k+2

4k = 14

k = 7/2  

Therefore answer the value of k is 7/2.

Answer 2

Let a = 3 K - 2


a+d = 4k -6


3k-2 +d = 4k -6


d = k-4


a3 = k+2


a+ 2d = k+2


3k-2 +2k -8= k+2


5k -10 = k+2


4k = 14


k = 7/2  



please subscribe to my channel➡️Ronit Antil

BeBrainly

@riskyjaaat