Question

Find the value of k For which 2k+7 , 6k-2, 8k+4 are 3 consecutive terms of an AP.​

Answer 1

Answer:

Step-by-step explanation:

Since the numbers are in AP,

2nd number - 1st number = 3rd number - 1st number

=> (6k - 2) - (2k + 7) = (8k + 4) - (6k - 2)

=> 4k - 9 = 2k + 6

=> 2k = 15

=> k = 15/2

So, the value of k is 15/2.

Hope it helps you my dear friend

Answer 2

$\huge\bf\blue{HEYA!}$

$<b> <i>$

HEre, 2k+7, 6k-2 and 8k+4. will be

consecutive terms of an AP, if.

2(6k-2)=(2k+7)+(8k+4)

=> 12k-4=10k+11

=> 12k-10k=11+4.

=> 2k=15

=> k=15/2.

HOPE THIS HELPS