Question

What Is The Value Of k Such That
$k + 1$
,
$3k - 1$
,
$4k + 1$
Are In A.P.


​

Answer 1

Answer:

k=4

Step-by-step explanation:

Let us assume that k+1,3k-1,4k+1........ are in AP

Therefore,

$a_{1} = k + 1$

$d = a_{2} - a_{1}$

and

$a_{2} - a_{1} = a_{3} - a_{2}$

$(3k - 1) - (k + 1) = (4k + 1) - (3k - 1)$

$3k - 1 - k - 1 = 4k + 1 - 3k + 1$

$2k - 2 = k + 2$

$2k - k = 2 + 2$

$k = 4$

Therefore the value of K is 4

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