Question

for what value of k,2k-1,k+5 and 3k+2 are three consecutive terms of Ap

Answer 1

If the common difference is constant, the terms are in AP.

Solution :-

T1 = 2k - 1

T2 = k + 5

T3 = 3k +2

Now,

T2 - T1 = T3 - T2

=> k + 5 - 2k + 1 = 3k +2 - k - 5

=> 6 - k = 2k - 3

=> 6 + 3 = 2k + k

=> 3k = 9

=> k = 3

Answer 2

Refer to the attachment!!
Hope this helps!!
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