If 2k,3k-1,8 are in A.P., then what is the value of k?

Open in App

Solution

$A_{1} = 2 K$
$A_{2} = 3 K - 1$
$A_{3} = 8$
$D = D$
$A_{2} - A_{1} = A_{3} - A_{2}$
$3 K - 1 - 2 K = 8 - (3 K - 1)$
$K - 1 = 8 - 3 K + 1$
$K + 3 K = 8 + 1 + 1$
$4 K = 10$
$K = \frac{10}{4}$
$K = 2.5$
we get, $K = 2.5$

Suggest Corrections

Similar questions

k : 2 k, k + 10

3 k + 2

A P

What value of k, 2k-1, 7 and 3K will form an AP.

If the points $(k, 2 k), (3 k, 3 k)$ and $(3, 1)$ are collinear then the value of $k$ is

k^{\circ}, k^{\circ}, 2 k^{\circ}, 2 k^{\circ}, and 3 k^{\circ}

k

(k, 2 k), (2 k, 3 k), (3, 1)

k

Join BYJU'S Learning Program