Question

Find the value of k for which 2k+7 ,6k-2 ,8k+4 are 3 consecutive terms of ap

Answer 1

2k+7+6k-2+8k+4=0
16k-9=0
16k=9
k=9/16

Answer 2

↑ Here is your answer ↓
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For an AP, common difference is the point

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

$6k - 2 - 2k - 7 = 8k + 4 - 6k + 2$

$4k - 9 = 2k+ 6$

Open now,

$4k - 9 - 2k - 6 = 0$

$2k - 15 = 0$

$2k = 15$

$k = 15/2$

$k = 7.5$

Now lets find the terms,

$2k+7 = 2(7.5) + 7 = 15 + 7 = 22$

$6k-2 = 6(7.5) - 2 = 45 -2 = 43$

$8k + 4 = 8(7.5) + 4 = 60+4 = 64$

We can see that in all cases, the common difference is 21 ^.^