If k-1, k+3 and 3k-1are in AP ;then find the value of k

1.	If k-1, k+3 and 3k-1are in AP ;then find the value of k
Answer» a3 -a2 = a2 -a1 3k-1 -(k+3) = k+3 -(K-1)3k-1 - k - 3 = k+3-k+12k -4 = 4 2k = 4+4 2k = 8 K = 8/2 K = 4 Send solution If it is in ap then d=(k+3)-(k-1)=> k+3-k+1=> 4(3k-1)-(k+3) should be 4=> 3k-1-k-3=4=> 2k-4=4=> 2k=8=> k = 4 Subtract 2nd term from 3rd term and also 1st term from 2nd term both will be equal as they arr commn diff. of an A.P. 4

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