Question

Find k, given that 3k+1, k and -3 are consecutive terms of an arithmetic sequence.

Answer 1

Consider

3k+1 = a

k = b

-3 = c

Now as all the terms are in arithmetic progression,

a,b,c

=> b-a = c-b

=> 2b = c+a

=> 2k = (-3)+(3k+1)

=> 2k = -3+3k+1

=> 2k = -2+3k

=> 2k-3k = -2

=> -k = -2

=> k = 2

Answer 2

Answer:

Hey dude here goes your answer

Step-by-step explanation:

Since 3k+1,k,-3 are consecutive terms of AP the difference is same

Therefore,   k-(3k+1)= -3-k

k-3k-1=  -3-k

-2k-1= -3-k

-k= -2

k=2