for what value of k 3k+2, 4k+5 and 6k-4 are in ap

Open in App

Solution

K, 3k+2,4k+5,6k-4 are in arithmetic progression
a2 - a1 = a3-a2
4k+5-(3k+2)=6k-4-(4k+5)
4k-3k+5-2=6k-4k-4-5
K+3=2k-9
-k=-9-3
K=12

Suggest Corrections

Similar questions

9 k - 6, 5 k - 4, 6 k - 17

k

2 k + 7, 6 k - 2

8 k + 4

3

(K + 2), (4 K - 6)

(3 K - 6)

Find the value of k if 8k + 4 = 6k - 2 = 2 k - 7 are three consecutive terms of an AP.

Determine k so that (3k -2), (4k-6) and (k +2) are three consecutive terms of an AP.

Join BYJU'S Learning Program